Logen-1-logen-1 - 4a - 1-n- 1-3n3-1-5n5- - - Find 8a-

log( 1+ 1 / n nbsp;) nbsp; log( 1 1 / n nbsp;) nbsp; =( 1 / n + 1 / 2n^ 2 + 1 / 3 n ^ 3 +...∞ nbsp;) ( 1 / n + 1 / 2 n ^ 2 1 / 3 ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

logen+1-logen...

Question

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

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Solution

$log (1 + \frac{1}{n}) - log (1 - \frac{1}{n}) = (\frac{1}{n} + \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} + . . . \infty) - (- \frac{1}{n} + \frac{1}{2 n^{2}} - \frac{1}{3 n^{3}} + . . . \infty)$
$\Rightarrow log (n + 1) - log (n - 1) = 2 (\frac{1}{n} + \frac{1}{3 n^{3}} + \frac{1}{5 n^{5}} + . . . \infty)$
Ans: 4

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loge(n+1)−loge(n−1)=4a[(1n)+(13n3)+(15n5)+...∞] Find 8a.

log(1+1n)−log(1−1n)=(1n+12n2+13n3+...∞)−(−1n+12n2−13n3+...∞)⇒log(n+1)−log(n−1)=2(1n+13n3+15n5+...∞)Ans: 4

$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .