1999 Binomial Identity Problem (2 Viewers)

frog1944 · Dec 3, 2016

Hi,

In the 1999 MX1 paper, question 7 B. How do you do it?

Thanks

He-Mann · Dec 3, 2016

(1-x)^n (1+ 1/x)^n = (whatever)^n, then maybe consider equating the coefficients of x^2?

Lugia101101 · Dec 4, 2016

$\begin{align*}\text{So }\left(1-x\right)^n\left(1+\frac{1}{x}\right)^n=\sum_{r=0}^{n}{{{n}\choose{r}}\left(-1\right)^rx^r}\sum_{k=0}^{n}{{{n}\choose{k}}x^{-k}}\end{align*}$
$\begin{align*}&\text{Now notice that if we let }x=1\text{ we get an expression with each term in the form }\pm{{n}\choose{a}}{{n}\choose{b}}\text{.}\\ &\text{To isolate the expression we want, we have to consider how the exponent of the }x\text{ affects the binomial coefficients before it.}\\ &\text{As we are dealing with the product of two sums, there will be many combinations of }r\text{ and }k\text{ that produce specific }x\text{ terms.}\\ &\text{Now note that the power of }x\text{ in the sum is }r-k\text{.}\\ &\text{We can see from the intermediate result given, that when the denominator of the choose functions are subtracted from one another, they equal 2.}\\ &\text{As the denominators in the sum are }r\text{ and }k\text{, and }r-k\text{ gives the exponent of }x\text{, we can deduce that }r-k=2\text{.}\\ &\text{So }0\leq r\leq n\text{, }0\leq k\leq n\text{, and }r-k=2\text{, which leaves }k=0,1,2,\cdots,(n-2)\text{ and }r=2,3,4,\cdots,n\text{ as the only valid values.}\end{align*}$
$\begin{align*}\text{Thus, the coefficient of }x^2\text{ is }&\sum_{r=2}^{n}{(-1)^r{{n}\choose{r}}{{n}\choose{r-2}}}\\ =&{{n}\choose{0}}{{n}\choose{2}}-{{n}\choose{1}}{{n}\choose{3}}+\cdots+(-1)^n{{n}\choose{n}}{{n}\choose{n-2}}\text{, which is what we want.}\end{align*}$
$\text{So we now know that the coefficient of }x^2\text{ in }\left(1-x\right)^n\left(1+\frac{1}{x}\right)^n\text{ is the result they want us to simplify.}$
$\begin{align*}\text{But we can also simplify }&\left(1-x\right)^n\left(1+\frac{1}{x}\right)^n\\ =&\left(1-x\right)^n\left(\frac{1+x}{x}\right)^n\\ =&\left(\frac{1-x^2}{x}\right)^n\\ =&\left(x^{-1}-x\right)^n\\ =&\sum_{p=0}^{n}{{{n}\choose{p}}(-1)^px^px^{-(n-p)}}\\ =&\sum_{p=0}^{n}{{{n}\choose{p}}(-1)^px^{2p-n}}\end{align*}$
$\begin{align*}\text{So the coefficient of }x^2\text{ in this occurs when }2p-n&=2\\ 2p&=2+n\\ p&=1+\frac{n}{2}\end{align*}$
$\begin{align*}&\text{So the coefficient of }x^2\text{ is }{{n}\choose{1+\frac{n}{2}}}(-1)^{1+\frac{n}{2}}\\ &\therefore{{n}\choose{0}}{{n}\choose{2}}-{{n}\choose{1}}{{n}\choose{3}}+\cdots+(-1)^n{{n}\choose{n}}{{n}\choose{n-2}}={{n}\choose{1+\frac{n}{2}}}(-1)^{1+\frac{n}{2}}\\ &\text{However, the choose function is only defined when the denominator is an integer, and }1+\frac{n}{2}\text{ is only an integer when }n\text{ is even.}\end{align*}$
$\begin{align*}\text{So, if }n\text{ is even, }&={{n}\choose{1+\frac{n}{2}}}(-1)^{1+\frac{n}{2}}\\ \text{If }n\text{ is odd, }&=0\text{, as the }x^2\text{ term will not exist and its coefficient will be }0\text{.}\end{align*}$

1999 Binomial Identity Problem (2 Viewers)

frog1944

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He-Mann

Vexed?

Lugia101101

New Member

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