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 know the property of binomial coefficients: $^{n}{C_{r}} = ^{n}{C_{n-r}}$.Thus,$^{n+r}{C_{n}} = ^{n+r}{C_{(n+r)-n}} = ^{n+r}{C_{r}}$.We need to

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know the property of binomial coefficients: $^{n}{C_{r}} = ^{n}{C_{n-r}}$.Thus,$^{n+r}{C_{n}} = ^{n+r}{C_{(n+r)-n}} = ^{n+r}{C_{r}}$.We need to evaluate the sum $S = \sum_{r=0}^{m} {^{n+r}{C_{r}}} = ^{n}{C_{0}} + ^{n+1}{C_{1}} + ^{n+2}{C_{2}} + \dots + ^{n+m}{C_{m}}$.Using the Hockey-stick identity,which states that $\sum_{i=r}^{n} {^{i}{C_{r}}} = ^{n+1}{C_{r+1}}$,we can rewrite our sum.Here,the sum is $\sum_{r=0}^{m} {^{n+r}{C_{r}}}$.By the identity $\sum_{i=0}^{m} {^{n+i}{C_{i}}} = ^{n+m+1}{C_{m}}$,or more standardly $\sum_{k=0}^{m} {^{n+k}{C_{k}}} = ^{n+m+1}{C_{m}}$.Since $^{n+m+1}{C_{m}} = ^{n+m+1}{C_{(n+m+1)-m}} = ^{n+m+1}{C_{n+1}}$.Therefore,the sum is equal to $^{n+m+1}{C_{n+1}}$.

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

Similar Questions

There are $6$ multiple choice questions in an examination. How many sequences of answers are possible,if the first three questions have $4$ choices each and the next three have $5$ choices each?

How many $3$-digit numbers are there with no digit repeated?

In a touring cricket team,there are $16$ players in all,including $5$ bowlers and $2$ wicket-keepers. How many teams of $11$ players can be chosen from these,such that each team includes exactly $3$ bowlers and $1$ wicket-keeper?

In how many ways can $7$ men and $7$ women be seated around a round table such that no two women sit together?

Vedclass Test Series

Exam Paper Generator

Online Exam Module