frac{{{C_0}}}{1} + frac{{{C_2}}}{3} + frac{{{ | vedclass

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

Question

(c) Putting the values of ${C_0},{C_2},{C_4}....,$we get $ = 1 + \frac{{n(n - 1)}}{{3.2!}} + \frac{{n(n - 1)(n - 2)(n - 3)}}{{5.4!}} + ....$

=$\fra

Vedclass · Accepted Answer

$\frac{{{2^n}}}{{n + 1}}$

Vedclass · Answer

(c) Putting the values of ${C_0},{C_2},{C_4}....,$we get $ = 1 + \frac{{n(n - 1)}}{{3.2!}} + \frac{{n(n - 1)(n - 2)(n - 3)}}{{5.4!}} + ....$

=$\frac{1}{{n + 1}}\left[ {(n + 1) + \frac{{(n + 1)n(n - 1)}}{{3!}} + \frac{{(n + 1)n(n - 1)(n - 2)(n - 3)}}{{5!}} + ....} \right]$

Put $n + 1$=N = $\frac{1}{N}\left[ {N + \frac{{N(N - 1)(N - 2)}}{{3!}} + \frac{{N(N - 1)\,(N - 2)(N - 3)(N - 4)}}{{5!}} + ....} \right]$

$ = \frac{1}{N}\left\{ {{\,^N}{C_1} + {\,^N}{C_3} + {\,^N}{C_5} + ....} \right\}$

$ = \frac{1}{N}\left\{ {{2^{N - 1}}} \right\} = \frac{{{2^n}}}{{n + 1}}$   $\{ N = n + 1\} $

Trick : Put $n=1$, then ${S_1} = \frac{{^1{C_0}}}{1} = \frac{1}{1} = 1$

At $n=2$, ${S_2} = \frac{{^2{C_0}}}{1} + \frac{{^2{C_2}}}{3} = 1 + \frac{1}{3} = \frac{4}{3}$

Also $(c)$ $ \Rightarrow \,\,\,{S_1} = 1,{S_2} = \frac{4}{3}$

$\frac{{{C_0}}}{1} + \frac{{{C_2}}}{3} + \frac{{{C_4}}}{5} + \frac{{{C_6}}}{7} + ....$=

Similar Questions

If ${C_0},{C_1},{C_2},.......,{C_n}$ are the binomial coefficients, then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals

The sum, of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$, is equal to

If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =

$\frac{{{C_0}}}{1} + \frac{{{C_1}}}{2} + \frac{{{C_2}}}{3} + .... + \frac{{{C_n}}}{{n + 1}} = $