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6.Permutation and Combination
hard
$\sum\limits_{r = 0}^m {^{n + r}{C_n} = } $
A
$^{n + m + 1}{C_{n + 1}}$
B
$^{n + m + 2}{C_n}$
C
$^{n + m + 3}{C_{n - 1}}$
D
इनमें से कोई नहीं
Solution
चूँकि $^n{C_r}{ = ^n}{C_{n – r}}$ तथा $^n{C_{r – 1}}{ + ^n}{C_r}{ = ^{n + 1}}{C_r}$
$\therefore $ $\sum\limits_{r = 0}^m {^{n + r}{C_n}} = \sum\limits_{r = 0}^m {^{n + r}{C_r}} { = ^n}{C_0}{ + ^{n + 1}}{C_1}{ + ^{n + 2}}{C_2} + ….{ + ^{n + m}}{C_m}$
$ = [1 + (n + 1)]{ + ^{n + 2}}{C_2}{ + ^{n + 3}}{C_3} + ……..{ + ^{n + m}}{C_m}$
${ = ^{n + m + 1}}{C_{n + 1}}$, $[\because {\;^n}{C_r}{ = ^n}{C_{n – r}}]$
Standard 11
Mathematics
