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

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(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

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

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} = } $

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