The expansion of 12-12-22-13-23-14-24- is

The correct option is C log_e(32) The expansion for log(1+x) is log( 1+x nbsp;)= nbsp;x x^22+ x^33 x^44+x^55 .... Therefore on substitutin ...

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Standard XII

Mathematics

Progression

The expansion...

Question

The expansion of $\frac{1}{2} - \frac{1}{{2.2}^{2}} + \frac{1}{{3.2}^{3}} - \frac{1}{{4.2}^{4}} \dots . .$ is

\frac{1}{4}

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{log}_{e} (\frac{3}{4})

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{log}_{e} (\frac{3}{2})

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{log}_{e} (\frac{2}{3})

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Solution

The correct option is C ${log}_{e} (\frac{3}{2})$
The expansion for $log (1 + x)$ is
$log (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} - . . . .$
Therefore on substituting $x = \frac{1}{2}$
We get ${log}_{e} (\frac{3}{2})$

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a = 1, b = {log}_{e} 2, c = {log}_{e} 3

, then which condition is true?

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Find the general value of ${l o g}_{3}$ (3i). Where i= $\sqrt{- 1}$

Q. 1.2 + 2.2² + 3.2³ + ... + n.2ⁿ= (n − 1) 2ⁿ⁺¹+2

$1.2 + {2.2}^{2} + {3.2}^{3} + \dots + n . 2^{n} = (n - 1) 2^{n + 1} + 2.$

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The expansion of 12−12.22+13.23−14.24….. is

The correct option is C loge(32)The expansion for log(1+x) islog(1+x)=x−x22+x33−x44+x55−....Therefore on substituting x=12 We get loge(32)

The expansion of $\frac{1}{2} - \frac{1}{{2.2}^{2}} + \frac{1}{{3.2}^{3}} - \frac{1}{{4.2}^{4}} \dots . .$ is

The correct option is C ${log}_{e} (\frac{3}{2})$
The expansion for $log (1 + x)$ is
$log (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} - . . . .$
Therefore on substituting $x = \frac{1}{2}$
We get ${log}_{e} (\frac{3}{2})$