frac{C_0}{1} + frac{C_1}{2} + frac{C_2}{3} + | vedclass

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

Question

(C) We know that $\frac{C_r}{r + 1} = \frac{1}{n + 1} \binom{n + 1}{r + 1}$.Substituting this into the given expression:$\sum_{r=0}^{n} \frac{C_r}{r +

Vedclass · Accepted Answer

$\frac{2^{n + 1} - 1}{n + 1}$

Vedclass · Answer

(C) We know that $\frac{C_r}{r + 1} = \frac{1}{n + 1} \binom{n + 1}{r + 1}$.Substituting this into the given expression:$\sum_{r=0}^{n} \frac{C_r}{r + 1} = \sum_{r=0}^{n} \frac{1}{n + 1} \binom{n + 1}{r + 1}$$= \frac{1}{n + 1} \sum_{r=0}^{n} \binom{n + 1}{r + 1}$Let $k = r + 1$. As $r$ goes from $0$ to $n$,$k$ goes from $1$ to $n + 1$.$= \frac{1}{n + 1} \sum_{k=1}^{n + 1} \binom{n + 1}{k}$Since $\sum_{k=0}^{m} \binom{m}{k} = 2^m$,we have $\sum_{k=1}^{m} \binom{m}{k} = 2^m - \binom{m}{0} = 2^m - 1$.Here $m = n + 1$,so the sum is $2^{n + 1} - 1$.Therefore,the expression equals $\frac{2^{n + 1} - 1}{n + 1}$.

$\frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + .... + \frac{C_n}{n + 1} = $

Similar Questions

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

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

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

If $(1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$,then $C_0 + 2 C_1 + 3 C_2 + \ldots + (n+1) C_n$ is equal to

Let $(1 + x)^m = C_0 + C_1x + C_2x^2 + C_3x^3 + . . . + C_mx^m$,where $C_r = {}^mC_r$ and $A = C_1C_3 + C_2C_4 + C_3C_5 + . . . + C_{m-2}C_m$. Which of the following is false?

Vedclass Test Series

Exam Paper Generator

Online Exam Module