Question

If `( 1 + x )^n = C_0 + C_1 x_1 + C_2 x_2 + ⋯ + C_n x_n` then show that `C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!) ?`

Answer 1

Please see below.

Explanation:

We know that

`(a+b)^n=C_0^na^n+C_1^na^(n-1)b+C_2^na^(n-2)b^2+ ⋯ + C_n^nb^n`

and `r^(th)` term is `C_r^na^rb^(n-r)` and `C_r^n=(n!)/(r!(n-r)!)`

and hence `(1+x)^n=C_0 1^n+C_1 1^(n-1)*x+C_2 1^(n-2)x^2+ ⋯ + C_nx^n`

= `(1+x)^n=C_0 1^n+C_1x+C_2x^2+ ⋯ + C_nx^n` (A)

and similarly `(x+1)^n=C_0 x^n+C_1 x^(n-1)*1+C_2 x^(n-2)1^2+ ⋯ + C_n1^n`

= `(x+1)^n=C_0x^n+C_1x^(n-1)+C_2x^(n-2)+ ⋯ +C_n1` (B)

Multipllying (A) and (B), we get `(1+x)^n(x+1)^n` on the LHS and multiplication of expansion on the RHS.

Now what is coefficient of `x^(n+r)` in this.

While on LHS we have `(1+x)^(2n)` and coefficient of `x^(n+r)` is `(2n!)/((n+r)!(2n-n-r)!)=(2n!)/((n+r)!(n-r)!)`

On the RHS we get `x^(n+r)`, when `C_0x^n` is multiplied by `C_rx^r`; `C_1x^(n-1)` is multiplied by `C_(r+1)x^(r+1)`; `C_2x^(n-2)` is multiplied by `C_(r+2)x^(r+2)` and so on till we get `C_nx^n` multipled by `C_(r+n)x^(n-(r+n)`.

Hence coefficient of `x^(n+r)` is

`C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)`

and hence `C_0C_r+C_1C_(r+1)+C_2C_(r+2)+....C_nC_(r+n)=((2n)!)/((n+r)!(n-r)!)`