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

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Question

(a) Since $^n{C_r}{ = ^n}{C_{n - r}}$ and $^n{C_{r - 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}$

we have $\sum\limits_{r = 0}^m {^{n + r}{C_n}} = \sum\limit

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Since $^n{C_r}{ = ^n}{C_{n - r}}$ and $^n{C_{r - 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}$

we have $\sum\limits_{r = 0}^m {^{n + r}{C_n}} = \sum\limits_{r = 0}^m {^{n + r}{C_r}} { = ^n}{C_0}{ + ^{n + 1}}{C_1}{ + ^{n + 2}}{C_2} + ......{ + ^{n + m}}{C_m}$

$ = [1 + (n + 1)]{ + ^{n + 2}}{C_2}{ + ^{n + 3}}{C_3} + ........{ + ^{n + m}}{C_m}$

${ = ^{n + m + 1}}{C_{n + 1}}$                $[\because {\;^n}{C_r}{ = ^n}{C_{n - r}}]$

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

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