7.Binomial Theorem
hard

$\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ का मान है

A

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

B

$n{.2^n}$

C

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

D

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

Solution

हम जानते हैं कि

$\frac{{{{(1 + x)}^n} – {{(1 – x)}^n}}}{2} = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + ….$

$x = 0$ से $x = 1$ तक समाकलन करने पर,

$\frac{1}{2}\int\limits_0^1 {\{ {{(1 + x)}^n} – {{(1 – x)}^n}\} \,} dx$

$ = \int\limits_0^1 {({C_1}x + {C_3}{x^3} + {C_5}{x^5} + ….)} dx$

$\frac{1}{2}\left\{ {\frac{{{{(1 + x)}^{n + 1}}}}{{n + 1}} + \frac{{{{(1 – x)}^{n + 1}}}}{{n + 1}}} \right\}_0^1 = \frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + ….$

या $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + …. = \frac{1}{2}\left\{ {\frac{{{2^{n + 1}} – 1}}{{n + 1}} + \frac{{0 – 1}}{{n + 1}}} \right\}$

 $ = \frac{1}{2}\left( {\frac{{{2^{n + 1}} – 2}}{{n + 1}}} \right) = \frac{{{2^n} – 1}}{{n + 1}}$

Standard 11
Mathematics

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