The value of C 1 2 + C 3 4 + C 5 6 + ...... is equal to Options: A . &nbsp 2 n − 1 n + 1 B . &nbsp n. 2 n C . &nbsp 2 n n D . &nbsp 2 n − 1

Answer: Option A : A We know that ( 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 1 2 ∫ 1 0 ( 1 + x ) n − ( 1 − x ) n d x = ∫ 1 0 ( C 1 x + C 3 x 3 + C 5 x 5 + . . . . . . . ) d x ⇒ 1 2 { ( 1 + x ) n + 1 n + 1 + ( 1 − x ) n + 1 n + 1 } 1 0 = C 1 2 + C 3 4 + C 5 6 + .... or C 1 2 + C 3 4 + C 5 6 + .......= 1 2 { 2 n + 1 − 1 n + 1 + 0 − 1 n + 1 } = 1 2 2 n + 1 − 2 n + 1 = 2 n − 1 n + 1

The value of C12 + C34 + C56 + ...... is equal to &#x

Discussion Forum : Binomial Theorem

Question -

The value of $\frac{C_{1}}{2}$ + $\frac{C_{3}}{4}$ + $\frac{C_{5}}{6}$ + ...... is equal to

Options:

A .

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

B . n.

2^{n}

C .

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

D .

2^{n - 1}

Answer: Option A
:
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

$\frac{1}{2}$ $\int_{0}^{1} (1 + x)^{n} - (1 - x)^{n} d x$

= $\int_{0}^{1} (C_{1} x + C_{3} x^{3} + C_{5} x^{5} + . . . . . . .) d x$

⇒ $\frac{1}{2} {\frac{(1 + x)^{n + 1}}{n + 1} + \frac{(1 - x)^{n + 1}}{n + 1}}_{0}^{1} = \frac{C_{1}}{2}$ + $\frac{C_{3}}{4}$ + $\frac{C_{5}}{6}$ + ....

or $\frac{C_{1}}{2}$ + $\frac{C_{3}}{4}$ + $\frac{C_{5}}{6}$ + .......= $\frac{1}{2}$ { $\frac{2^{n + 1} - 1}{n + 1}$ + $\frac{0 - 1}{n + 1}$ }

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

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