C-0-2-C-1-3-C-2-4-C-3-5-

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} + \dots \dots \dots .

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} d x = \frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots

f o r t h e i n t e g r a l l e t 1 - x = t

\frac{C_{0}}{2} - \frac{C_{1}}{3} + \frac{C_{2}}{4} - \dots = - \int_{1}^{0} (1 - t) t^{n} d t

= \int_{0}^{1} (t^{n} - t^{n + 1}) d t

= (\frac{t^{n + 1}}{n + 1} - \frac{t^{n + 2}}{n + 2}) ∣_{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

Thank you very much Sir!