frac{{{C_1}}}{2} + frac{{{C_3}}}{4} + frac{{{ | vedclass

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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}}$

$\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ =. .. .

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