The value of frac{{{C_1}}}{2} + frac{{{C_3}}} | vedclass

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

Question

(a) We know that

$\frac{{{{(1 + x)}^n} - {{(1 - x)}^n}}}{2} = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + ....$

Integrating from $x = 0$ to $x = 1$, we g

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) We know that

$\frac{{{{(1 + x)}^n} - {{(1 - x)}^n}}}{2} = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + ....$

Integrating from $x = 0$ to $x = 1$, we get$\frac{1}{2}\int\limits_0^1 {\{ {{(1 + x)}^n} - {{(1 - x)}^n}\} \,} dx$

$ = \int\limits_0^1 {({C_1}x + {C_3}{x^3} + {C_5}{x^5} + ....)} dx$

==> $\frac{1}{2}\left\{ {\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}} + \frac{{{{(1 - x)}^{n + 1}}}}{{n + 1}}} \right\}_0^1 = \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + ....$

or $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .... = \frac{1}{2}\left\{ {\frac{{{2^{n + 1}} - 1}}{{n + 1}} + \frac{{0 - 1}}{{n + 1}}} \right\}$

$ = \frac{1}{2}\left( {\frac{{{2^{n + 1}} - 2}}{{n + 1}}} \right) = \frac{{{2^n} - 1}}{{n + 1}}$

The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ is equal to

Similar Questions

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

The sum of last eigth coefficients in the expansion of $(1 + x)^{15}$ is :-

If ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_{15}}{x^{15}},$ then ${C_2} + 2{C_3} + 3{C_4} + .... + 14{C_{15}} = $

Let $n$ be an odd integer. If $\sin n\theta = \sum\limits_{r = 0}^n {{b_r}{{\sin }^r}\theta } $ for every value of $\theta $, then

If ${\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3}\, = \frac{k}{{21}}$, then $k$ equals