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

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free