Binomial Theorem 3Unit Australia Year 12 If X is the sum of the first, third, fifth, ... terms in the expansion of $$ (a+b)^n $$ and Y is the sum of the second, fourth, sixth, ... terms, show that $$ X^2 - Y^2 = (a^2-b^2)^n $$
So if I expand it, it is basically the sum of the odd terms would include:
$$ X=\binom{n}{0} +\binom{n}{2}x^2 + \binom{n}{4}x^4 + \binom{n}{6}x^6+ ... $$
and the sum of the even terms would include
$$ Y= \binom{n}{1}x +\binom{n}{3}x^3 + \binom{n}{5}x^5 + \binom{n}{7}x^7+ ... $$
What would you do like factorise (X-Y)(X+Y) or something but theres like infinite numbers n

The terms of the expansion of $(a-b)^n$ are the same as those of $(a+b)^n$ except that the terms for odd powers of $b$ are multiplied by $-1$.  Thus
$$ X + Y = (a+b)^n$$
while
$$ X - Y = (a-b)^n$$
So $$ X^2 - Y^2 = (X + Y)(X - Y) = (a+b)^n (a-b)^n = (a^2-b^2)^n$$
