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So I was solving problems in my textbook and I got stuck on this question. It says:

Prove that:

$ ^nC_0^2 - ^nC_1^2 + ^nC_2^2 - ^nC_3^2 + ... + (-1)^{n} .^nC_n^2 $ = $ 0 $ or $ \frac {(-1)^{n/2} n!} {\frac{n}{2}! \frac{n}{2}!} $ ,

when $ n $ is odd/even.

So I tried to bring the required terms and my idea was to multiply $ (1+x)^n $ with $(x - 1)^n$ .

$$ (1+x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + ... ^nC_n x^n $$ $$ (x-1)^n = ^nC_0x^n - ^nC_1 x^{n-1} + ^nC_2 x^{n-2} + ... (-1)^n .^nC_n $$

I can see that if I multiply the two equations, then I will get the terms I need with their required sign. LHS will contain $ (x^2 - 1)^n $ .

But along with that, there will be many other terms on the RHS which will probably be impossible for me to get rid of. Am I going in the right way or is there an entirely different procedure?

Manish Kundu
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1 Answers1

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I think your $C$s are missing an index. Presuming that your sum is ${n\choose 0}^2-{n\choose 1}^2+{n\choose2}^2-\ldots$ $=\sum_{j=0}^n(-1)^j{n\choose j}^2$, then you're basically on the right track; I'd keep in mind that ${n\choose j} = {n\choose n-j}$ (which you seem to be doing implicitly) and write out the product that you have. Note that your sum will be exactly the $[x^n]$ term of this result - can you see why that would be so? Once you know that, you're almost home, because if two expressions are equal as polynomials in $x$ then they'll be equal termwise, so you know that the $[x^n]$ term of $(1+x)^n(x-1)^n$ will be the same as the $x^n$ term of $(x^2-1)^n$.

Steven Stadnicki
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  • Yeah so $ x^n $ of the expansion is my required sum – Manish Kundu Jan 06 '18 at 09:51
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    exactly. And that must be equal to the $x^n$ term on $(x^2-1)^n$, which you should be able to show is equal to the given answer - in other words, if $n$ is odd then this term is zero (why?) and if $n=2j$, say, it's $(-1)^j{2j\choose j}$. – Steven Stadnicki Jan 06 '18 at 09:54
  • I get that term as $ ^n C_\frac {n}{2} x^n (-1)^{n/2} $ . My answer comes if n is even, and for odd, $ ^n C_\frac {n}{2} $ does not exist. So I can directly conclude that it is zero, right? – Manish Kundu Jan 06 '18 at 10:08
  • Not exactly - it's more accurate to say that when $n$ is odd, the $[x^n]$ term of $(x^2-1)^n$ doesn't exist because $(x^2-1)^n$ is an even function and so only has even powers of $x$ in its expansion. – Steven Stadnicki Jan 06 '18 at 12:00
  • Yeah thanks. Answer accepted. – Manish Kundu Jan 06 '18 at 12:46