frac{1}{{1!(n - 1),!}} + frac{1}{{3!(n - 3)!} | vedclass

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

Question

(b) Multiplying each term by $n!$ the question reduces to

$\frac{{n!}}{{1!(n - 1)!}} + \frac{1}{{3!}}.\frac{{n!}}{{(n - 3)\,!}} + \frac{1}{{5!}}.\f

Vedclass · Accepted Answer

$\frac{{{2^{n - 1}}}}{{n!}}$; for all values of $n$ i.e., all even odd values

Vedclass · Answer

(b) Multiplying each term by $n!$ the question reduces to

$\frac{{n!}}{{1!(n - 1)!}} + \frac{1}{{3!}}.\frac{{n!}}{{(n - 3)\,!}} + \frac{1}{{5!}}.\frac{{n!}}{{(n - 5)!}} + ....$

$ = {\,^n}{C_1} + {\,^n}{C_3} + {\,^n}{C_5} + .... = {2^{n - 1}}$.

Thus $\frac{1}{{1!(n - 1)!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + ....$$ = \frac{1}{{n!}}{2^{n - 1}}$.

$\frac{1}{{1!(n - 1)\,!}} + \frac{1}{{3!(n - 3)!}} + \frac{1}{{5!(n - 5)!}} + .... = $

Similar Questions

The value of $^{15}C_0^2{ - ^{15}}C_1^2{ + ^{15}}C_2^2 - ....{ - ^{15}}C_{15}^2$ is

The sum of the coefficients in the expansion of ${(1 + x - 3{x^2})^{2163}}$ will be

If $S_n =$$\sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and $T_n =$$\sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $ then $\frac{{{T_n}}}{{{S_n}}}$ is equal to

The coefficient of $x^{4}$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{6}$ in powers of $x,$ is

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