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Question Number 113354 by Dwaipayan Shikari last updated on 12/Sep/20

$$\frac{\left(\begin{array}{c}n\\ 0\end{array}\right)}{2}-\frac{\left(\begin{array}{c}n\\ 1\end{array}\right)}{3}+\frac{\left(\begin{array}{c}n\\ 2\end{array}\right)}{4}-\frac{\left(\begin{array}{c}n\\ 3\end{array}\right)}{5}+.....n$$

Commented by mr W last updated on 13/Sep/20

$${(1-x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{(-1)}^k{x}^k$$$$x^2{(1-x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{(-1)}^k{x}^{k+2}$$$${\int }_0^1{x}^2{(1-x)}^{n}dx=\underset{k=0}{\sum ^n}{C}_k^n{(-1)}^k{\underset{0}{\int }}^1{x}^{k+2}dx$$$${\int }_0^1{x}^2{(1-x)}^{n}dx=\underset{k=0}{\sum ^n}{(-1)}^k\frac{C_k^n}{k+2}$$$$...$$

Commented by maths mind last updated on 13/Sep/20

$$nicesir\beta (a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}={\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dxcanhelpyou$$$$$$$$toofinishihopyoudoingwellverrynicedaysir$$

Commented by mr W last updated on 14/Sep/20

$$thanksalotsir!$$

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