The sum of the series 1-loge x1-loge x22- is

The correct option is A x We've, 1+log_e x1!+(log_e x)^22!+........ =e^log_ex [ Since e^x=1+x1!+x^22!+........ ] =x . ...

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Standard XII

Mathematics

Summation by Sigma Method

The sum of th...

Question

The sum of the series $1 + \frac{{log}_{e} x}{1!} + \frac{({log}_{e} x)^{2}}{2!} + . . . . . . . .$ is

x

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x^{2}

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x^{3}

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none of these

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Solution

The correct option is A $x$
We've,
$1 + \frac{{log}_{e} x}{1!} + \frac{({log}_{e} x)^{2}}{2!} + . . . . . . . .$
$= e^{{log}_{e} x}$ [ Since $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . . . . . .$ ]
$= x$ .

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The sum of the series 1+logex1!+(logex)22!+........ is

The correct option is A xWe've,1+logex1!+(logex)22!+........=elogex[ Since ex=1+x1!+x22!+........]=x.

The sum of the series $1 + \frac{{log}_{e} x}{1!} + \frac{({log}_{e} x)^{2}}{2!} + . . . . . . . .$ is

The correct option is A $x$
We've,
$1 + \frac{{log}_{e} x}{1!} + \frac{({log}_{e} x)^{2}}{2!} + . . . . . . . .$
$= e^{{log}_{e} x}$ [ Since $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + . . . . . . . .$ ]
$= x$ .