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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
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-\index Index
-\shortcut idx
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc1[10]
-\end_layout
-
-\end_inset
-
-The text says that 
-\begin_inset Formula $a_{1}+a_{2}+\dots+a_{0}=0$
-\end_inset
-
-.
- What then,
- is 
-\begin_inset Formula $a_{2}+\dots+a_{0}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We could set that to be 
-\begin_inset Formula $-a_{1}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc2[01]
-\end_layout
-
-\end_inset
-
-What does the notation 
-\begin_inset Formula $\sum_{1\leq j\leq n}a_{j}$
-\end_inset
-
- mean,
- if 
-\begin_inset Formula $n=3.14$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $a_{1}+a_{2}+a_{3}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc3[13]
-\end_layout
-
-\end_inset
-
-Without using the 
-\begin_inset Formula $\sum$
-\end_inset
-
--notation,
- write out the equivalent of
-\begin_inset Formula 
-\[
-\sum_{0\leq n\leq5}\frac{1}{2n+1},
-\]
-
-\end_inset
-
-and also the equivalent of
-\begin_inset Formula 
-\[
-\sum_{0\leq n^{2}\leq5}\frac{1}{2n^{2}+1}.
-\]
-
-\end_inset
-
-Explain why the two results are different,
- in spite of rule (b).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{eqnarray*}
-\sum_{0\leq n\leq5}\frac{1}{2n+1} & = & \frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11},\\
-\sum_{0\leq n^{2}\leq5}\frac{1}{2n^{2}+1} & = & \frac{1}{9}+\frac{1}{3}+\frac{1}{1}+\frac{1}{3}+\frac{1}{9},
-\end{eqnarray*}
-
-\end_inset
-
-as the integers with 
-\begin_inset Formula $0\leq n^{2}\leq5$
-\end_inset
-
- are 
-\begin_inset Formula $\{-2,-1,0,1,2\}$
-\end_inset
-
-.
- Rule (b) states that a permutation of the integers satisfying the condition doesn't alter the result,
- but because 
-\begin_inset Formula $n\mapsto n^{2}$
-\end_inset
-
- is not a permutation of such 
-\emph on
-integers
-\emph default
-,
- the rule doesn't apply.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc4[10]
-\end_layout
-
-\end_inset
-
-Without using the 
-\begin_inset Formula $\sum$
-\end_inset
-
--notation,
- write out the equivalent of each side of Eq.
- (10) as a sum of sums for the case 
-\begin_inset Formula $n=3$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $a_{11}+a_{21}+a_{22}+a_{31}+a_{32}+a_{33}=a_{11}+a_{21}+a_{31}+a_{22}+a_{32}+a_{33}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc5[HM20]
-\end_layout
-
-\end_inset
-
-Prove that rule (a) is valid for arbitrary infinite series,
- provided that the series converge.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Since infinite series are defined for cases where the total number of nonzero elements is at most countable,
- we might assume that the series are indexed by positive integers.
- Thus,
- we'd have to prove that
-\begin_inset Formula 
-\[
-\left(\sum_{i=1}^{\infty}a_{i}\right)\left(\sum_{j=1}^{\infty}b_{j}\right)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{i}b_{j},
-\]
-
-\end_inset
-
-assuming that the relevant series converge.
- We have
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\[
-\left(\sum_{i=0}^{\infty}a_{i}\right)\left(\sum_{j=0}^{\infty}b_{j}\right)=\sum_{i=0}^{\infty}\left(a_{i}\sum_{j=0}^{\infty}b_{j}\right)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}b_{j},
-\]
-
-\end_inset
-
-by entering the second series as a constant into the first series (because it converges) and then entering the element of the first series,
- considered as a constant,
- into the inner series.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc9[05]
-\end_layout
-
-\end_inset
-
-Is the derivation of Eq.
- (14) valid even if 
-\begin_inset Formula $n=-1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The result is
-\begin_inset Formula 
-\[
-\sum_{0\leq j\leq-1}ax^{j}=0=a\left(\frac{1-x^{0}}{1-x}\right),
-\]
-
-\end_inset
-
-which is correct.
- However,
- the derivation is not correct as the rule (d) cannot be applied,
- because the applications in the derivation assume 
-\begin_inset Formula $n\geq0$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc10[05]
-\end_layout
-
-\end_inset
-
-Is the derivation of Eq.
- (14) valid even if 
-\begin_inset Formula $n=-2$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-No (see previous exercise).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc11[03]
-\end_layout
-
-\end_inset
-
-What should the right-hand side of Eq.
- (14) be if 
-\begin_inset Formula $x=1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Mere substitution in the original formula yields an undefined result,
- but it's clear that
-\begin_inset Formula 
-\[
-\sum_{0\leq j\leq n}a\cdot1^{j}=\sum_{0\leq j\leq n}a=(n+1)a.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc12[10]
-\end_layout
-
-\end_inset
-
-What is 
-\begin_inset Formula $1+\frac{1}{7}+\frac{1}{49}+\frac{1}{343}+\dots+\left(\frac{1}{7}\right)^{n}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-By Eq.
- (14),
- this is
-\begin_inset Formula 
-\[
-\sum_{k=0}^{n}\left(\frac{1}{7}\right)^{k}=\left(\frac{1-\left(\frac{1}{7}\right)^{n+1}}{1-\frac{1}{7}}\right)=\frac{1-\left(\frac{1}{7}\right)^{n+1}}{\frac{6}{7}}=\frac{7}{6}\left(1-\frac{1}{7^{n+1}}\right).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc13[10]
-\end_layout
-
-\end_inset
-
-Using Eq.
- (15) and assuming that 
-\begin_inset Formula $m\leq n$
-\end_inset
-
-,
- evaluate 
-\begin_inset Formula $\sum_{j=m}^{n}j$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{multline*}
-\sum_{m\leq j\leq n}j=\sum_{m\leq j+m\leq n}(j+m)=\sum_{0\leq j\leq n-m}(m+j)=\\
-=m(n-m+1)+\frac{1}{2}(n-m)(n-m+1)=\frac{1}{2}(n(n+1)-m(m-1)).
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc14[11]
-\end_layout
-
-\end_inset
-
-Using the result of the previous exercise,
- evaluate 
-\begin_inset Formula $\sum_{j=m}^{n}\sum_{k=r}^{s}jk$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{multline*}
-\sum_{j=m}^{n}\sum_{k=r}^{s}jk=\left(\sum_{j=m}^{n}j\right)\left(\sum_{k=r}^{s}k\right)=\\
-=\frac{1}{4}(n(n+1)-m(m-1))(s(s+1)-r(r-1)).
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc15[M22]
-\end_layout
-
-\end_inset
-
-Compute the sum 
-\begin_inset Formula $1\times2+2\times2^{2}+3\times2^{3}+\dots+n\times2^{n}$
-\end_inset
-
- for small values of 
-\begin_inset Formula $n$
-\end_inset
-
-.
- Do you see the pattern developing in these numbers?
- If not,
- discover it by manipulations similar to those leading up to Eq.
- (14).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $\tau(n)$
-\end_inset
-
- be such a such sum
-\begin_inset Formula 
-\begin{align*}
-\tau(1) & =2, & \tau(2) & =2+8=10, & \tau(3) & =10+24=34,\\
-\tau(4) & =34+64=98, & \tau(5) & =98+160=258, &  & \dots
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-Now,
-\begin_inset Formula 
-\begin{align*}
-\tau(n) & =\sum_{1\leq k\leq n}k2^{k} &  & \text{by definition}\\
- & =\sum_{0\leq k\leq n-1}(k+1)2^{k+1} &  & \text{by rule (b)}\\
- & =2\sum_{0\leq k\leq n-1}(k+1)2^{k} &  & \text{by a special case of (a)}\\
- & =2\left(\sum_{0\leq k\leq n-1}k2^{k}+\sum_{0\leq k\leq n-1}2^{k}\right) &  & \text{by rule (c)}\\
- & =2\left(\sum_{0\leq k\leq n}k2^{k}-n2^{n}+(2^{n}-1)\right) &  & \text{by rule (d) and Eq. (14)}\\
- & =2\left(\sum_{1\leq k\leq n}k2^{k}-n2^{n}+(2^{n}-1)\right) &  & \text{by rule (d)}
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-This means 
-\begin_inset Formula $\tau(n)=2(\tau(n)-n2^{n}+2^{n}-1)=2\tau(n)-n2^{n+1}+2^{n+1}-2$
-\end_inset
-
-,
- so 
-\begin_inset Formula 
-\[
-\tau(n)=n2^{n+1}-2^{n+1}+2=(n-1)2^{n+1}+2.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc17[M00]
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $S$
-\end_inset
-
- be a set of integers.
- What is 
-\begin_inset Formula $\sum_{j\in S}1$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula $|S|$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc20[25]
-\end_layout
-
-\end_inset
-
-Dr.
- I.
- J.
- Matrix has observed a remarkable sequence of formulas:
-\begin_inset Formula 
-\begin{align*}
-9\times1+2 & =11, & 9\times12+3 & =111,\\
-9\times123+4 & =1111, & 9\times1234+5 & =11111.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Write the good doctor's great discovery in terms of the 
-\begin_inset Formula $\sum$
-\end_inset
-
--notation.
-\end_layout
-
-\begin_layout Enumerate
-Your answer to part (a) undoubtedly involves the number 10 as base of the decimal system;
- generalize this formula so that you get a formula that will perhaps work in any base 
-\begin_inset Formula $b$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Prove your formula from part (b) by using formulas derived in the text or in exercise 16 above.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-For 
-\begin_inset Formula $n\geq1$
-\end_inset
-
-,
- 
-\begin_inset Formula 
-\[
-9\sum_{j=0}^{n-1}10^{j}(n-j)+n+1=\sum_{j=0}^{n}10^{j}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula 
-\[
-(b-1)\sum_{j=0}^{n-1}b^{j}(n-j)+n+1=\sum_{j=0}^{n}b^{j}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-We have
-\begin_inset Formula 
-\begin{eqnarray*}
-\sum_{j=0}^{n-1}b^{j}(n-j) & = & n\sum_{j=0}^{n-1}b^{j}-\sum_{j=0}^{n-1}jb^{j}\\
- & = & n\frac{1-b^{n}}{1-b}-\frac{(n-1)b^{n+1}-nb^{n}+b}{(b-1)^{2}}\\
- & = & n\frac{b^{n}-1}{b-1}-\frac{(n-1)b^{n+1}-nb^{n}+b}{(b-1)^{2}},
-\end{eqnarray*}
-
-\end_inset
-
-so
-\begin_inset Formula 
-\begin{align*}
- & (b-1)\sum_{j=0}^{n-1}b^{j}(n-j)+n+1\\
- & =n(b^{n}-1)-\frac{(n-1)b^{n+1}-nb^{n}+b}{b-1}+n+1\\
- & =\frac{n(b^{n}-1)(b-1)-(n-1)b^{n+1}+nb^{n}-b+n(b-1)+b-1}{b-1}\\
- & =\frac{nb^{n+1}-nb^{n}-nb+n-nb^{n+1}+b^{n+1}+nb^{n}-b+nb-n+b-1}{b-1}\\
- & =\frac{b^{n+1}-1}{b-1}=\frac{1-b^{n+1}}{b-1}=\sum_{j=0}^{n}b^{j}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc21[M25]
-\end_layout
-
-\end_inset
-
-Derive rule (d) from (8) and (17).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{multline*}
-\sum_{R(j)}a_{j}+\sum_{S(j)}a_{j}=\sum_{j}a_{j}[R(j)]+\sum_{j}a_{j}[S(j)]=\sum_{j}a_{j}([R(j)]+[S(j)])=\\
-\sum_{j}a_{j}([R(j)\lor S(j)]+[R(j)\land S(j)])=\sum_{R(j)\lor S(j)}a_{j}+\sum_{R(j)\land S(j)}a_{j}.
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc22[20]
-\end_layout
-
-\end_inset
-
-State the appropriate analogs of Eqs.
- (5),
- (7),
- (8),
- and (11) for 
-\emph on
-products
-\emph default
- instead of sums.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{align*}
-\prod_{R(i)}a_{i} & =\prod_{R(j)}a_{j}=\prod_{R(p(j))}a_{p(j)}, & \prod_{R(i)}\prod_{S(j)}a_{ij} & =\prod_{S(j)}\prod_{R(i)}a_{ij},\\
-\prod_{R(i)}b_{i}c_{i} & =\prod_{R(i)}b_{i}+\prod_{R(i)}c_{i}, & \prod_{R(j)}a_{j}\prod_{S(j)}a_{j} & =\prod_{R(j)\lor S(j)}a_{j}\prod_{R(j)\land S(j)}a_{j}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc23[10]
-\end_layout
-
-\end_inset
-
-Explain why it is a good idea to define 
-\begin_inset Formula $\sum_{R(j)}a_{j}$
-\end_inset
-
- and 
-\begin_inset Formula $\prod_{R(j)}a_{j}$
-\end_inset
-
- as zero and one,
- respectively,
- when no integers satisfy 
-\begin_inset Formula $R(j)$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Zero and one are the neutral terms for sum and product,
- respectively,
- so by defining it this way,
- rule (d) (among others) applies independently of whether there are integers satisfying the properties or not.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc25[15]
-\end_layout
-
-\end_inset
-
-Consider the following derivation;
- is anything amiss?
-\begin_inset Formula 
-\[
-\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{j=1}^{n}\frac{1}{a_{j}}\right)=\sum_{1\leq i\leq n}\sum_{1\leq j\leq n}\frac{a_{i}}{a_{j}}=\sum_{1\leq i\leq n}\sum_{1\leq i\leq n}\frac{a_{i}}{a_{i}}=\sum_{i=1}^{n}1=n.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-First,
- the change of variable on the second equality is invalid since it changes to a bound variable,
- not to a free variable.
- Second,
- it then converts the double summation into a single summation,
- which is invalid too.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc29[M30]
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-Express 
-\begin_inset Formula $\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=0}^{j}a_{i}a_{j}a_{k}$
-\end_inset
-
- in terms of the multiple-sum notation explained at the end of the section.
-\end_layout
-
-\begin_layout Enumerate
-Express the same sum in terms of 
-\begin_inset Formula $\sum_{i=0}^{n}a_{i}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\sum_{i=0}^{n}a_{i}^{2}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $\sum_{i=0}^{n}a_{i}^{3}$
-\end_inset
-
- [see Eq.
- (13)].
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula 
-\[
-\sum_{0\leq k\leq j\leq i\leq n}a_{i}a_{j}a_{k}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_deeper
-\begin_layout Standard
-\noindent
-We have
-\begin_inset Formula 
-\begin{align*}
-S & :=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=0}^{j}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=j}^{i}a_{i}a_{j}a_{k}\\
- & =\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=0}^{i}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=0}^{i}\sum_{k=i}^{n}a_{i}a_{j}a_{k}\\
- & =\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=i}^{j}a_{i}a_{j}a_{k}=\sum_{i=0}^{n}\sum_{j=i}^{n}\sum_{k=j}^{n}a_{i}a_{j}a_{k}.
-\end{align*}
-
-\end_inset
-
-Thus,
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-bgroup
-\backslash
-small
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{eqnarray*}
-6S & = & \sum_{i=0}^{n}\sum_{j=0}^{i}\left(\sum_{k=0}^{j}a_{i}a_{j}a_{k}+\sum_{k=j}^{i}a_{i}a_{j}a_{k}+\sum_{k=i}^{n}a_{i}a_{j}a_{k}\right)\\
- &  & +\sum_{i=0}^{n}\sum_{j=i}^{n}\left(\sum_{k=0}^{i}a_{i}a_{j}a_{k}+\sum_{k=i}^{j}a_{i}a_{j}a_{k}+\sum_{i=j}^{n}a_{i}a_{j}a_{k}\right)\\
- & = & \sum_{i=0}^{n}\left(\sum_{j=0}^{i}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{i}^{2}a_{j}\right)+\sum_{j=i}^{n}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{j}^{2}a_{i}\right)\right)\\
- & = & \sum_{i=0}^{n}\left(\sum_{j=0}^{n}\left(\sum_{k=0}^{n}a_{i}a_{j}a_{k}+a_{i}a_{j}^{2}+a_{i}^{2}a_{j}\right)+\sum_{k=0}^{n}a_{i}^{2}a_{k}+a_{i}^{3}+a_{i}^{3}\right)\\
- & = & \sum_{i=0}^{n}\sum_{j=0}^{n}\sum_{k=0}^{n}a_{i}a_{j}a_{k}+\sum_{i=0}^{n}\sum_{j=0}^{n}a_{i}a_{j}^{2}+\sum_{i=0}^{n}\sum_{j=0}^{n}a_{i}^{2}a_{j}+\sum_{i=0}^{n}\sum_{k=0}^{n}a_{i}^{2}a_{k}+2\sum_{i=0}^{n}a_{i}^{3}\\
- & = & \left(\sum_{i=0}^{n}a_{i}\right)^{3}+\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+\left(\sum_{i=0}^{n}a_{i}^{2}\right)\left(\sum_{i=0}^{n}a_{i}\right)\\
- &  & +\left(\sum_{i=0}^{n}a_{i}^{2}\right)\left(\sum_{i=0}^{n}a_{i}\right)+2\sum_{i=0}^{n}a_{i}^{3}\\
- & = & \left(\sum_{i=0}^{n}a_{i}\right)^{3}+3\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+2\left(\sum_{i=0}^{n}a_{i}^{3}\right).
-\end{eqnarray*}
-
-\end_inset
-
-
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-egroup 
-\end_layout
-
-\end_inset
-
-This means
-\begin_inset Formula 
-\[
-S=\frac{1}{6}\left(\sum_{i=0}^{n}a_{i}\right)^{3}+\frac{1}{2}\left(\sum_{i=0}^{n}a_{i}\right)\left(\sum_{i=0}^{n}a_{i}^{2}\right)+\frac{1}{3}\left(\sum_{i=0}^{n}a_{i}^{3}\right).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_deeper
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc30[M23]
-\end_layout
-
-\end_inset
-
-(J.
- Binet,
- 1812.) Without using induction,
- prove the identity 
-\begin_inset Formula 
-\[
-\left(\sum_{j=1}^{n}a_{j}x_{j}\right)\left(\sum_{j=1}^{n}b_{j}y_{j}\right)=\left(\sum_{j=1}^{n}a_{j}y_{j}\right)\left(\sum_{j=1}^{n}b_{j}x_{j}\right)+\sum_{1\leq j<k\leq n}(a_{j}b_{k}-a_{k}b_{j})(x_{j}y_{k}-x_{k}y_{j}).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\begin{eqnarray*}
-\left(\sum_{j=1}^{n}a_{j}x_{j}\right)\left(\sum_{j=1}^{n}b_{j}y_{j}\right) & = & \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}x_{i}y_{j}\\
- & = & \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}(x_{j}y_{i}-x_{j}y_{i}+x_{i}y_{j})\\
- & = & \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}x_{j}y_{i}+\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i})\\
- & = & \left(\sum_{j=1}^{n}a_{j}y_{j}\right)\left(\sum_{j=1}^{n}b_{j}x_{j}\right)+\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i}),
-\end{eqnarray*}
-
-\end_inset
-
-but
-\begin_inset Formula 
-\begin{eqnarray*}
-\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i}) & = & \sum_{1\leq j<i\leq n}a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i})+\sum_{1\leq i<j\leq n}a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i})\\
- & = & \sum_{1\leq i<j\leq n}(a_{j}b_{i}(x_{j}y_{i}-x_{i}y_{j})+a_{i}b_{j}(x_{i}y_{j}-x_{j}y_{i}))\\
- & = & \sum_{1\leq j<k\leq n}(a_{j}b_{k}-a_{k}b_{j})(x_{j}y_{k}-x_{k}y_{j}).
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset Note Comment
-status open
-
-\begin_layout Plain Layout
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc33[M30]
-\end_layout
-
-\end_inset
-
-One evening Dr.
- Matrix discovered some formulas that might even be classed as more remarkable than those of exercise 20:
-\end_layout
-
-\begin_layout Plain Layout
-\begin_inset Formula 
-\begin{align*}
-\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)} & =0,\\
-\frac{a}{(a-b)(a-c)}+\frac{b}{(b-a)(b-c)}+\frac{c}{(c-a)(c-b)} & =0,\\
-\frac{a^{2}}{(a-b)(a-c)}+\frac{b^{2}}{(b-a)(b-c)}+\frac{c^{2}}{(c-a)(c-b)} & =1,\\
-\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)} & =a+b+c.
-\end{align*}
-
-\end_inset
-
-Prove that these formulas are a special case of a general law;
- let 
-\begin_inset Formula $x_{1},x_{2},\dots,x_{n}$
-\end_inset
-
- be distinct numbers,
- and show that
-\begin_inset Formula 
-\[
-\sum_{j=1}^{n}\Bigg(x_{j}^{r}\Bigg/\prod_{\begin{subarray}{c}
-1\leq k\leq n\\
-k\neq j
-\end{subarray}}(x_{j}-x_{k})\Bigg)=\begin{cases}
-0, & \text{if }0\leq r<n-1;\\
-1, & \text{if }r=n-1;\\
-{\textstyle \sum_{j=1}^{n}x_{j}}, & \text{if }r=n.
-\end{cases}
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Plain Layout
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc37[M24]
-\end_layout
-
-\end_inset
-
-Show that the determinant of Vandermonde's matrix is
-\begin_inset Formula 
-\[
-\prod_{1\leq j\leq n}x_{j}\prod_{1\leq i<j\leq n}(x_{j}-x_{i}).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We show this by induction.
- For 
-\begin_inset Formula $n=1$
-\end_inset
-
-,
- this is obvious.
- For 
-\begin_inset Formula $n>1$
-\end_inset
-
-,
- assuming this holds for 
-\begin_inset Formula $n-1$
-\end_inset
-
-,
- subtracting from each row the previous one multiplied by 
-\begin_inset Formula $x_{1}$
-\end_inset
-
-,
- expanding by cofactors,
- factoring out,
- and applying the induction hypothesis,
- we have
-\begin_inset Formula 
-\begin{eqnarray*}
-\left|\begin{array}{cccc}
-x_{1} & x_{2} & \cdots & x_{n}\\
-x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2}\\
-\vdots & \vdots &  & \vdots\\
-x_{1}^{n} & x_{2}^{n} & \cdots & x_{n}^{n}
-\end{array}\right| & = & \left|\begin{array}{cccc}
-x_{1} & x_{2} & \cdots & x_{n}\\
-0 & x_{2}(x_{2}-x_{1}) & \cdots & x_{n}(x_{n}-x_{1})\\
-\vdots & \vdots &  & \vdots\\
-0 & x_{2}^{n-1}(x_{2}-x_{1}) & \cdots & x_{n}^{n-1}(x_{n}-x_{1})
-\end{array}\right|\\
- & = & x_{1}\left|\begin{array}{ccc}
-x_{2}(x_{2}-x_{1}) & \cdots & x_{n}(x_{n}-x_{1})\\
-\vdots &  & \vdots\\
-x_{2}^{n-1}(x_{2}-x_{1}) & \cdots & x_{n}^{n-1}(x_{n}-x_{1})
-\end{array}\right|\\
- & = & x_{1}(x_{2}-x_{1})\cdots(x_{n}-x_{1})\left|\begin{array}{ccc}
-x_{2} & \cdots & x_{n}\\
-\vdots &  & \vdots\\
-x_{2}^{n-1} & \cdots & x_{n}^{n-1}
-\end{array}\right|\\
- & = & x_{1}\prod_{2\leq j\leq n}(x_{j}-x_{1})\prod_{2\leq j\leq n}x_{j}\prod_{2\leq i<j\leq n}(x_{j}-x_{i})\\
- & = & \prod_{1\leq j\leq n}x_{j}\prod_{1\leq i<j\leq n}(x_{j}-x_{i}).
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc38[M25]
-\end_layout
-
-\end_inset
-
-Show that the determinant of Cauchy's matrix is
-\begin_inset Formula 
-\[
-\prod_{1\leq i<j\leq n}(x_{j}-x_{i})(y_{j}-y_{i})\Bigg/\prod_{1\leq i,j\leq n}(x_{i}+y_{j}).
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We prove it by induction.
- For 
-\begin_inset Formula $n=1$
-\end_inset
-
-,
- this is obvious.
- Now assume 
-\begin_inset Formula $n>1$
-\end_inset
-
- and that the hypothesis holds for 
-\begin_inset Formula $n-1$
-\end_inset
-
-.
- Multiplying each row 
-\begin_inset Formula $i$
-\end_inset
-
- by 
-\begin_inset Formula $\frac{x_{1}+y_{1}}{x_{i}+y_{1}}$
-\end_inset
-
- times the first row,
- we get
-\begin_inset Formula 
-\begin{multline*}
-\left|\begin{array}{cccc}
-\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\
-\frac{1}{x_{2}+y_{1}} & \frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\
-\vdots & \vdots &  & \vdots\\
-\frac{1}{x_{n}+y_{1}} & \frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}}
-\end{array}\right|=\\
-=\left|\begin{array}{cccc}
-\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\
-0 & \frac{1}{x_{2}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{2}+y_{1})}\\
-\vdots & \vdots &  & \vdots\\
-0 & \frac{1}{x_{n}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{n}+y_{1})}
-\end{array}\right|=\\
-=\frac{1}{x_{1}+y_{1}}\left|\begin{array}{ccc}
-\frac{1}{x_{2}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{2}+y_{1})}\\
-\vdots &  & \vdots\\
-\frac{1}{x_{n}+y_{2}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{n})(x_{n}+y_{1})}
-\end{array}\right|.
-\end{multline*}
-
-\end_inset
-
-Now,
-\begin_inset Formula 
-\begin{multline*}
-\frac{1}{x_{i}+y_{j}}-\frac{x_{1}+y_{1}}{(x_{1}+y_{j})(x_{i}+y_{1})}=\frac{(x_{1}+y_{j})(x_{i}+y_{1})-(x_{1}+y_{1})(x_{i}+y_{j})}{(x_{i}+y_{j})(x_{1}+y_{j})(x_{i}+y_{1})}=\\
-=\frac{1}{x_{i}+y_{j}}\frac{x_{1}y_{1}+x_{i}y_{j}-x_{1}y_{j}-x_{i}y_{1}}{(x_{1}+y_{j})(x_{i}+y_{1})}=\frac{1}{x_{i}+y_{j}}\frac{(x_{i}-x_{1})(y_{j}-y_{1})}{(x_{1}+y_{j})(x_{i}+y_{1})},
-\end{multline*}
-
-\end_inset
-
-so
-\begin_inset Formula 
-\begin{multline*}
-\left|\begin{array}{cccc}
-\frac{1}{x_{1}+y_{1}} & \frac{1}{x_{1}+y_{2}} & \cdots & \frac{1}{x_{1}+y_{n}}\\
-\frac{1}{x_{2}+y_{1}} & \frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\
-\vdots & \vdots &  & \vdots\\
-\frac{1}{x_{n}+y_{1}} & \frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}}
-\end{array}\right|=\\
-=\frac{1}{x_{1}+y_{1}}\left|\begin{array}{ccc}
-\frac{1}{x_{2}+y_{2}}\frac{(x_{2}-x_{1})(y_{2}-y_{1})}{(x_{1}+y_{2})(x_{2}+y_{1})} & \cdots & \frac{1}{x_{2}+y_{n}}\frac{(x_{2}-x_{1})(y_{n}-y_{1})}{(x_{1}+y_{n})(x_{2}+y_{1})}\\
-\vdots &  & \vdots\\
-\frac{1}{x_{n}+y_{2}}\frac{(x_{n}-x_{1})(y_{2}-y_{1})}{(x_{1}+y_{2})(x_{n}+y_{1})} & \cdots & \frac{1}{x_{n}+y_{n}}\frac{(x_{n}-x_{1})(y_{n}-y_{1})}{(x_{1}+y_{n})(x_{n}+y_{1})}
-\end{array}\right|=\\
-=\frac{1}{x_{1}+y_{1}}\prod_{i=2}^{n}\frac{x_{i}-x_{1}}{x_{i}+y_{1}}\prod_{j=2}^{n}\frac{y_{j}-y_{1}}{x_{1}+y_{j}}\left|\begin{array}{ccc}
-\frac{1}{x_{2}+y_{2}} & \cdots & \frac{1}{x_{2}+y_{n}}\\
-\vdots &  & \vdots\\
-\frac{1}{x_{n}+y_{2}} & \cdots & \frac{1}{x_{n}+y_{n}}
-\end{array}\right|=\\
-=\frac{\prod_{j=2}^{n}(x_{j}-x_{1})(y_{j}-y_{1})}{(x_{1}+y_{1})\prod_{i=2}^{n}(x_{j}+y_{1})\prod_{j=2}^{n}(x_{1}+y_{j})}\frac{\prod_{2\leq i<j\leq n}(x_{j}-x_{i})(y_{j}-y_{i})}{\prod_{2\leq i,j\leq n}(x_{i}+y_{j})}=\\
-=\prod_{1\leq i<j\leq n}(x_{j}-x_{i})(y_{j}-y_{i})\Bigg/\prod_{1\leq i,j\leq n}(x_{i}+y_{j}).
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc45[M25]
-\end_layout
-
-\end_inset
-
-A 
-\emph on
-Hilbert matrix
-\emph default
-,
- sometimes called an 
-\begin_inset Formula $n\times n$
-\end_inset
-
- segment of 
-\emph on
-the
-\emph default
- (infinite) Hilbert matrix,
- is a matrix for which 
-\begin_inset Formula $a_{ij}=1/(i+j-1)$
-\end_inset
-
-.
- Show that this is a special case of Cauchy's matrix,
- find its inverse,
- show that each element of the inverse is an integer and show that the sum of all elements of the inverse is 
-\begin_inset Formula $n^{2}$
-\end_inset
-
-.
- The solution to this problem requires an elementary knowledge of factorial and binomial coefficients,
- which are discussed in sections 1.2.5 and 1.2.6.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Clearly this is the Cauchy's matrix 
-\begin_inset Formula $a_{ij}=1/(x_{i}+y_{j})$
-\end_inset
-
- where 
-\begin_inset Formula $x_{i}=i$
-\end_inset
-
- and 
-\begin_inset Formula $y_{j}=j-1$
-\end_inset
-
-.
- Then the elements of the inverse,
- by exercise 41,
- are given by
-\begin_inset Formula 
-\begin{align*}
-b_{ij} & =\left(\prod_{1\leq k\leq n}(x_{j}+y_{k})(x_{k}+y_{i})\right)\Bigg/(x_{j}+y_{i})\Bigg(\prod_{\begin{subarray}{c}
-1\leq k\leq n\\
-k\neq j
-\end{subarray}}(x_{j}-x_{k})\Bigg)\Bigg(\prod_{\begin{subarray}{c}
-1\leq k\leq n\\
-k\neq i
-\end{subarray}}(y_{i}-y_{k})\Bigg)\\
- & =\left(\prod_{k}(j+k-1)(i+k-1)\right)\Bigg/(i+j-1)\Bigg(\prod_{k\neq j}(j-k)\Bigg)\Bigg(\prod_{k\neq i}(i-k)\Bigg)\\
- & =\frac{(j+n-1)!(i+n-1)!}{(j-1)!(i-1)!}\bigg/(i+j-1)(-1)^{i+j}(n-j)!(j-1)!(n-i)!(i-1)!\\
- & =(-1)^{i+j}\frac{(j+n-1)!(i+n-1)!}{(i+j-1)(n-j)!(n-i)!(j-1)!^{2}(i-1)!^{2}}.
-\end{align*}
-
-\end_inset
-
-We have used that
-\begin_inset Formula 
-\begin{multline*}
-\prod_{k\neq j}(j-k)=(-1)^{n}\prod_{k\neq j}(k-j)=(-1)^{n}\prod_{j<k\leq n}(k-j)\prod_{1\leq k<j}(k-j)=\\
-=(-1)^{n+j}(n-j)!(j-1)!,
-\end{multline*}
-
-\end_inset
-
-and the same happens to 
-\begin_inset Formula $i$
-\end_inset
-
-.
- Then,
-\begin_inset Formula 
-\begin{align*}
-b_{ij} & =\frac{(-1)^{i+j}ij}{i+j-1}\frac{(i+n-1)!(j+n-1)!}{i!(i-1)!(n-i)!j!(j-1)!(n-j)!}\\
- & =\frac{(-1)^{i+j}ij}{i+j-1}\binom{n}{i}\frac{(i+n-1)!}{n!(i-1)!}\binom{n}{j}\frac{(j+n-1)!}{n!}\\
- & =\frac{(-1)^{i+j}ij}{i+j-1}\binom{n}{i}\binom{i+n-1}{n}\binom{n}{j}\binom{j+n-1}{n}\\
- & =\frac{(-1)^{i+j}ij}{i+j-1}\binom{n}{i}\binom{-i}{n}\binom{n}{j}\binom{-j}{n}\\
- & =\frac{(-1)^{i+j}ij}{i+j-1}\binom{-j}{i}\binom{-j-i}{n-i}\binom{-i}{n}\binom{n}{j}\\
- & =\frac{(-1)^{i+j}ij}{i+j-1}\binom{i+j-1}{i}\binom{n+j-1}{n-i}\binom{n+i-1}{n}\binom{n}{j}\\
- & =(-1)^{i+j}\binom{i+j-2}{i-1}\binom{i+n-1}{i-1}\binom{j+n-1}{n-1}\binom{n}{j}\in\mathbb{Z}.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-Finally,
- from exercise 44,
-\begin_inset Formula 
-\[
-\sum_{i,j}b_{ij}=\sum_{i=1}^{n}i+\sum_{j=1}^{n}(j-1)=\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=n^{2}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document