The value of sumlimits_{n = 1}^infty {frac{{^ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}}} +

Vedclass · Accepted Answer

${e^2} - 1$

Vedclass · Answer

(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$.

The value of $\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ...{ + ^n}{C_n}}}{{^n{P_n}}}} $ is

Similar Questions

In the expansion of ${(1 + x)^{50}},$ the sum of the coefficient of odd powers of $x$ is

Given $(1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + ....$ and that $a_1^2\,= 2a_2$ then the value of $n$ is

The value of $-{ }^{15} C _{1}+2 .{ }^{15} C _{2}-3 .{ }^{15} C _{3}+\ldots \ldots$ $-15 .{ }^{15} C _{15}+{ }^{14} C _{1}+{ }^{14} C _{3}+{ }^{14} C _{5}+\ldots .+{ }^{14} C _{11}$ is

The coefficient of $x^{49}$ in the expansion of $(x - 1)$$\left( {x\, - \,\frac{1}{2}\,} \right)$$\left( {x\, - \,\frac{1}{{{2^2}}}\,} \right)$ .....$\left( {x\, - \,\frac{1}{{{2^{49}}}}\,} \right)$ is equal to

$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to