The n th term in the expansion of log e 4-3 isA- -1n-1-mB- -1nn-1-n1- 3n-1D- -1 n -1- n - 2 n

The correct option is B ( 1)^n 1n.3^nlog_e(43)=log_e(1+13) =13 12(13)^2+13(13)^3 ... Now n^th term of log_e(1+x) is =( 1)^n 1x^nn ∴ n^th term of the given seri ...

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Byju's Answer

Standard XII

Mathematics

Progression

The n th term...

Question

The $n^{th}$ term in the expansion of ${log}_{e} (\frac{4}{3})$ is

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

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\frac{(- 1)^{n - 1}}{n}

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

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\frac{(- 1)^{n - 1}}{n {.4}^{n}}

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Solution

The correct option is C $\frac{(- 1)^{n - 1}}{n {.3}^{n}}$
${log}_{e} (\frac{4}{3}) = {log}_{e} (1 + \frac{1}{3}) = \frac{1}{3} - \frac{1}{2} {(\frac{1}{3})}^{2} + \frac{1}{3} {(\frac{1}{3})}^{3} - . . .$
Now $n^{th}$ term of ${log}_{e} (1 + x)$ is $= (- 1)^{n - 1} \frac{x^{n}}{n}$
$∴ n^{th}$ term of the given series is
$= \frac{(- 1)^{n - 1}}{n} {(\frac{1}{3})}^{n}$

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The nth term in the expansion of loge(43) is

The correct option is C (−1)n−1n.3nloge(43)=loge(1+13)=13−12(13)2+13(13)3−... Now nth term of loge(1+x) is =(−1)n−1xnn ∴nth term of the given series is =(−1)n−1n(13)n

The $n^{th}$ term in the expansion of ${log}_{e} (\frac{4}{3})$ is