# 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)!) ?

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Feb 3, 2018

#### 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 ${r}^{t h}$ term is ${C}_{r}^{n} {a}^{r} {b}^{n - r}$ 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 ${\left(1 + x\right)}^{n} {\left(x + 1\right)}^{n}$ 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 ${\left(1 + x\right)}^{2 n}$ 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 ${C}_{0} {x}^{n}$ is multiplied by ${C}_{r} {x}^{r}$; ${C}_{1} {x}^{n - 1}$ is multiplied by ${C}_{r + 1} {x}^{r + 1}$; ${C}_{2} {x}^{n - 2}$ is multiplied by ${C}_{r + 2} {x}^{r + 2}$ and so on till we get ${C}_{n} {x}^{n}$ multipled by C_(r+n)x^(n-(r+n).

Hence coefficient of ${x}^{n + r}$ is

${C}_{0} {C}_{r} + {C}_{1} {C}_{r + 1} + {C}_{2} {C}_{r + 2} + \ldots . {C}_{n} {C}_{r + n}$

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

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