Logen-1-logen-1 - t-1-n - 1-3n3 - 1-5n5-Find t-

log_e(n+1) log_e(n 1) =log_e ( n+1n 1) =log _ e ( 1+ 1 n nbsp;/ 1 1 n nbsp; nbsp;) =log_e (1+1n) log_e (1 1n) = (1n 12n^2+ ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

logen+1-logen...

Question

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

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Solution

${log}_{e} (n + 1) - {log}_{e} (n - 1)$
$= {log}_{e} (\frac{n + 1}{n - 1})$
$= {log}_{e} ⎛ ⎜ ⎜ ⎝ \frac{1 + \frac{1}{n}}{1 - \frac{1}{n}} ⎞ ⎟ ⎟ ⎠$
$= {log}_{e} (1 + \frac{1}{n}) - {log}_{e} (1 - \frac{1}{n})$
$= (\frac{1}{n} - \frac{1}{2 n^{2}} + \frac{1}{3 n^{3}} - . . . . . . . . .) - (- \frac{1}{n} - \frac{1}{2 n^{2}} - \frac{1}{3 n^{3}} - . . . . . . . . .)$
$= 2 [\frac{1}{n} + \frac{1}{3 n^{3}} + \frac{1}{5 n^{5}} + . . . . . \infty]$
On comparing with given expression
$t = 2$

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

loge(n+1)−loge(n−1)=loge(n+1n−1)=loge⎛⎜ ⎜⎝1+1n1−1n⎞⎟ ⎟⎠=loge(1+1n)−loge(1−1n)=(1n−12n2+13n3−.........)−(−1n−12n2−13n3−.........)=2[1n+13n3+15n5+.....∞]On comparing with given expression t=2

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