Question

If `(1+x)^n=c_0+c_1x+c_2x^2+cdots+c_nx^n` then show that `c_0.c_n+c_1.c_(n-1)+c_2.c_(n-2)+cdots+c_n.c_0=((2n)!)/(n!)^2`?

Answer 1

Please see below.

Explanation:

As `(1+x)^n=c_0+c_1x+c_2x^2+......+c_nx^n`

consider coefficient of `x^n` in `(1+x)^nxx(1+x)^n`

this will be `c_0.c_n+c_1.c_(n-1)+c_2.c_(n-2)+......+c_n.c_0`

but `(1+x)^nxx(1+x)^n=(1+x)^(2n)` and coefficient of `x^n` in this is `c_n^(2n)=((2n)!)/(n!(2n-n)!)`

Hence `c_0.c_n+c_1.c_(n-1)+c_2.c_(n-2)+......+c_n.c_0=((2n)!)/(n!(2n-n)!)=((2n)!)/(n!)^2`