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

If ${C_r}$ stands for $^n{C_r}$,the sum of the series $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$,where $n$ is an even positive integer,is

${ }^{34}C_{10} + 3 \cdot { }^{34}C_{9} + 3 \cdot { }^{34}C_{8} + { }^{34}C_{7} = $

If $p$ and $q$ are positive integers,then the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ are

If $\sum_{r=0}^{20} {}^{20+r}C_r = \frac{p}{q} {}^{40}C_{20}$ and $GCD(p, q) = 1$,then $p^2 - q^2 =$

If $(1+x)^n = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n$,then the value of $p_0 + p_3 + p_6 + \ldots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module