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Question Number 22618 by Tinkutara last updated on 21/Oct/17

$$If{(1+x)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3$$$$+...+C_n{x}^n,provethat$$$$\frac{2^2}{1.2}{C}_0+\frac{2^3}{2.3}{C}_1+\frac{2^4}{3.4}{C}_2+...+$$$$\frac{2^{n+2}}{(n+1)(n+2)}{C}_n=\frac{3^{n+2}-2n-5}{(n+1)(n+2)}$$

Answered by ajfour last updated on 21/Oct/17

$${\int }_0^x{(1+x)}^{n}dx=\frac{C_{0}x}{1}+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+..$$$$\Rightarrow \frac{{(1+x)}^{n+1}-1}{n+1}=\frac{C_{0}x}{1}+\frac{C_1{x}^2}{2}+\frac{C_3{x}^3}{3}+...$$$${\int }_0^2\frac{{(1+x)}^{n+1}-1}{n+1}dx=\frac{2^2{C}_0}{1.2}+\frac{2^3{C}_1}{2.3}+..$$$$=\left[\frac{{(1+x)}^{n+2}-(n+2)x}{(n+1)(n+2)}\right]{\mid }_0^2$$$$=\frac{3^{n+2}-2n-4-1}{(n+1)(n+2)}=\frac{3^{\mathit{n}+2}-2\mathit{n}-5}{(\mathit{n}+1)(\mathit{n}+2)}.$$

Commented by Tinkutara last updated on 21/Oct/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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