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Question Number 22177 by Tinkutara last updated on 12/Oct/17

$$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+..........$$

Answered by ajfour last updated on 12/Oct/17

$$x{(1-x)}^n=C_{0}x-C_1{x}^2+C_2{x}^3+..$$$${\int }_0^{1}x{(1-x)}^{n}dx=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-...$$$$fortheintegrallet1-x=t$$$$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-...=-{\int }_1^0(1-t)t^{n}dt$$$$={\int }_0^1(t^n-t^{n+1})dt$$$$=(\frac{t^{n+1}}{n+1}-\frac{t^{n+2}}{n+2}){\mid }_0^1$$$$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}.$$

Commented by Tinkutara last updated on 13/Oct/17

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

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