_ k = 1 ^n k ( 1 + 1 n ) ^ k - 1 = - Vidyadip

MCQ

${\sum\limits_{k = 1}^n {k\left( {1 + \frac{1}{n}} \right)} ^{k - 1}} = $

A
$n(n - 1)$
B
$n(n + 1)$
✓
${n^2}$
D
${(n + 1)^2}$

✓

Answer

Correct option: C.

${n^2}$

c
(c)$\sum\limits_{k\, = 1}^n {k\,{{\left( {1 + \frac{1}{n}} \right)}^{k - 1}}} $

$ = 1 + 2{\left( {1 + \frac{1}{n}} \right)^1} + 3{\left( {1 + \frac{1}{n}} \right)^2} + ....$upto n terms

$=1 + 2t + 3{t^2} + ...$ upto $n$ terms

$={(1 - t)^{ - 2}} = {\left[ {1 - \left( {1 + \frac{1}{n}} \right)} \right]^{ - 2}}$ = ${\left( {\frac{1}{n}} \right)^{ - 2}} = {(n)^2} = {n^2}$.

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