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

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(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

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

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(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} = } $

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