The sum to (n + 1) terms of the series frac{C | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + (-1)^n C_n x^n$.Multiplying by $x$,we get $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 -

Vedclass · Accepted Answer

$\frac{1}{(n + 1)(n + 2)}$

Vedclass · Answer

(C) We know that $(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + \dots + (-1)^n C_n x^n$.Multiplying by $x$,we get $x(1 - x)^n = C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots + (-1)^n C_n x^{n+1}$.Integrating both sides from $0$ to $1$:$\int_0^1 x(1 - x)^n dx = \int_0^1 (C_0x - C_1x^2 + C_2x^3 - C_3x^4 + \dots) dx$.On the $R$.$H$.$S$.,we get $\left[ \frac{C_0x^2}{2} - \frac{C_1x^3}{3} + \frac{C_2x^4}{4} - \dots \right]_0^1 = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \dots$.On the $L$.$H$.$S$.,let $1 - x = t$,so $dx = -dt$. When $x=0, t=1$ and when $x=1, t=0$.$\int_1^0 (1 - t)t^n (-dt) = \int_0^1 (t^n - t^{n+1}) dt = \left[ \frac{t^{n+1}}{n+1} - \frac{t^{n+2}}{n+2} \right]_0^1 = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}$.Thus,the sum is $\frac{1}{(n+1)(n+2)}$.

The sum to $(n + 1)$ terms of the series $\frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \dots$ is

Similar Questions

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

If $(1-x+x^2)^n = a_0 + a_1 x + \ldots + a_{2n} x^{2n}$, then the value of $a_0 + a_2 + a_4 + \ldots + a_{2n}$ is

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n$,then the value of $C_0 + C_2 + C_4 + C_6 + ...$ is

The value of $\frac{1}{1! 50!} + \frac{1}{3! 48!} + \frac{1}{5! 46!} + \dots + \frac{1}{49! 2!} + \frac{1}{51! 1!}$ is $.............$.

If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module