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7.Binomial Theorem
hard
$\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ...{ + ^n}{C_n}}}{{^n{P_n}}}} $ = . . .
A
${e^2}$
B
$e$
C
${e^2} - 1$
D
$e - 1$
Solution
(c) $\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ……. + {\,^n}{C_n}}}{{{\,^n}{P_n}}}} $
$ = \frac{{{\,^1}{C_0} + {\,^1}{C_1}}}{{{\,^1}{P_1}}} + \frac{{{\,^2}{C_0} + {\,^2}{C_1} + {\,^2}{C_2}}}{{{\,^2}{P_2}}} + \frac{{^3{C_0} + {\,^3}{C_1} + {\,^3}{C_2} + {\,^3}{C_3}}}{{{\,^3}{P_3}}}$+…
$ = \frac{{{2^1}}}{{1!}} + \frac{{{2^2}}}{{2!}} + \frac{{{2^3}}}{{3!}} + …….$ $\left( {1 + \frac{2}{{1!}} + \frac{{{2^2}}}{{2!}} + \frac{{{2^3}}}{{3!}} + …….} \right) – 1$
$ = {e^2} – 1$.
Standard 11
Mathematics
