sumlimits_{r = 0}^m {^{n + r}{C_n} = } | vedclass

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

Question

(A) We use the identity $^{n}{C_{r}} = {^{n}}{C_{n-r}}$ and the Hockey-stick identity $\sum_{i=r}^{n} {^{i}{C_{r}}} = {^{n+1}}{C_{r+1}}$.Given the sum

Vedclass · Accepted Answer

$^{n + m + 1}{C_{n + 1}}$

Vedclass · Answer

(A) We use the identity $^{n}{C_{r}} = {^{n}}{C_{n-r}}$ and the Hockey-stick identity $\sum_{i=r}^{n} {^{i}{C_{r}}} = {^{n+1}}{C_{r+1}}$.Given the sum $S = \sum_{r=0}^{m} {^{n+r}{C_{n}}}$.Using the property $^{n+r}{C_{n}} = {^{n+r}}{C_{(n+r)-n}} = {^{n+r}}{C_{r}}$,we have:$S = \sum_{r=0}^{m} {^{n+r}{C_{r}}} = {^{n}{C_{0}}} + {^{n+1}}{C_{1}} + {^{n+2}}{C_{2}} + \dots + {^{n+m}}{C_{m}}$.By the Hockey-stick identity,$\sum_{k=0}^{m} {^{n+k}{C_{k}}} = {^{n+m+1}}{C_{m}}$.Since ${^{n+m+1}}{C_{m}} = {^{n+m+1}}{C_{(n+m+1)-m}} = {^{n+m+1}}{C_{n+1}}$,the result is ${^{n+m+1}}{C_{n+1}}$.

$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $

Similar Questions

If $n$ is a positive integer greater than $1$,then $3({ }^n C_0) - 8({ }^n C_1) + 13({ }^n C_2) - 18({ }^n C_3) + \ldots$ up to $(n+1)$ terms $=$

Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $,then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

Choose the correct option regarding the following statements:
$1$. $C_0+C_2+C_4+\ldots+C_n=2^{n-1}$,if $n$ is even
$2$. $C_1+C_3+C_5+\ldots+C_{n-1}=2^{n-1}$,if $n$ is even

If the sum of the coefficients of even powers of $x$ in the expansion of $(1-x+x^2)^{2n}$ is $3281$,then $n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module