sumlimits_{n = 1}^infty {frac{{^n{C_0} | vedclass

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Question

$\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}}}

Vedclass · Accepted Answer

${e^2} - 1$

Vedclass · Answer

$\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$.

$\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ...{ + ^n}{C_n}}}{{^n{P_n}}}} $ का मान है

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