The correct option is A 2^n 1n + 1 (1 = x)^n = ^nC_0 + ^nC_1X.... Integrate nbsp; (1 + x)^n + 1n + 1 = ^nC_0x + ^nC_1x^22 ... First with limi ...

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Combination

C12 + C34 + C...

Question

$\frac{C_{1}}{2} + \frac{C_{3}}{4} + \frac{C_{5}}{6} + . . .$

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

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

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

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\frac{1}{n + 1}

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Solution

The correct option is A $\frac{2^{n} - 1}{n + 1}$
$(1 = x)^{n} =^{n} C_{0} +^{n} C_{1} X . . . .$
Integrate
$\frac{(1 + x)^{n + 1}}{n + 1} =^{n} C_{0} x + \frac{^{n} C_{1} x^{2}}{2} . . .$
First with limits 0 to 1
$\frac{2^{m + 1}}{m + 1} - \frac{1}{m + 1} =^{n} C_{0} + \frac{^{n} C_{1}}{2} . . . (1)$
then -1 to 0
$\frac{(1)^{m + 1}}{m + 1} = 0 = 0 -$
$^{n} C_{0} (- 1) + \frac{^{n} C_{1} x^{2}}{2}$
$\frac{1}{n + 1} =^{n} C_{0} - \frac{^{n} C_{1}}{2} . . . . . . . . . (2)$
Subtracting (1) & (2)
$\frac{2^{n} - 1}{n + 1} =^{n} C_{1} 2 + \frac{^{n} C_{3}}{4}$

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Similar questions

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

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

C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + \frac{C_{3}}{4} + . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{(n + 1)}

1 - 2 n + \frac{2 n (2 n - 1)}{2!} - \frac{2 n (2 n - 1) (2 n - 1)}{3!} + . . . + (- 1)^{n - 1} \frac{2 n (n - 1) . . . (n + 2)}{(n - 1)}

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{9} - . . . + \frac{(- 1)^{n} C_{n}}{n^{2} + 2 n + 1}

= \frac{1}{n + 1} + \frac{1}{2 n + 2} + . . . + \frac{1}{n^{2} + 2 n + 1} .

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