1 2 .2 2 1^3 + 2 2 .3 2 1^3 + 2^3 + 3 2 .4 2 1^3 + 2^3 + 3^3 + ..… - Vidyadip

MCQ

$\frac{{\frac{1}{2}.\frac{2}{2}}}{{{1^3}}} + \frac{{\frac{2}{2}.\frac{3}{2}}}{{{1^3} + {2^3}}} + \frac{{\frac{3}{2}.\frac{4}{2}}}{{{1^3} + {2^3} + {3^3}}} + .....n$ terms =

A
${\left( {\frac{n}{{n + 1}}} \right)^2}$
B
${\left( {\frac{n}{{n + 1}}} \right)^3}$
✓
$\left( {\frac{n}{{n + 1}}} \right)$
D
$\left( {\frac{1}{{n + 1}}} \right)$

✓

Answer

Correct option: C.

$\left( {\frac{n}{{n + 1}}} \right)$

c
(c) ${T_n} = \frac{{\frac{{n(n + 1)}}{{2.\,2}}}}{{{1^3} + {2^3} + {3^3} + ..... + {n^3}}}$

$= \frac{{\frac{{n(n + 1)}}{4}}}{{{{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}}} $

$= \frac{1}{{n(n + 1)}} = \frac{1}{n} - \frac{1}{{n + 1}}$

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

$ = \left( {1 - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ....... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$

$ = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

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