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

I f {(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3}

+ \dots + C_{n} x^{n}, p r o v e t h a t

\frac{2^{2}}{1.2} C_{0} + \frac{2^{3}}{2.3} C_{1} + \frac{2^{4}}{3.4} C_{2} + \dots +

\frac{2^{n + 2}}{(n + 1) (n + 2)} C_{n} = \frac{3^{n + 2} - 2 n - 5}{(n + 1) (n + 2)}

Answered by ajfour last updated on 21/Oct/17

\int_{0}^{x} {(1 + x)}^{n} d x = \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} + \dots

\int_{0}^{2} \frac{{(1 + x)}^{n + 1} - 1}{n + 1} d x = \frac{2^{2} C_{0}}{1.2} + \frac{2^{3} C_{1}}{2.3} + . .

= [\frac{{(1 + x)}^{n + 2} - (n + 2) x}{(n + 1) (n + 2)}] ∣_{0}^{2}

= \frac{3^{n + 2} - 2 n - 4 - 1}{(n + 1) (n + 2)} = \frac{3^{n + 2} - 2 n - 5}{(n + 1) (n + 2)} .

Commented by Tinkutara last updated on 21/Oct/17

Thank you very much Sir!