A) \[\frac{1}{2}\]
B) 1
C) 2
D) 4
Correct Answer: C
Solution :
Given, \[{{S}_{n}}=\frac{1}{{{1}^{3}}}+\frac{1+2}{{{1}^{3}}+{{2}^{3}}}+....\] \[+\frac{1+2+3+...+n}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+....+{{n}^{3}}}\] Now, \[{{T}_{n}}=\frac{1+2+3+...+n}{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+....+{{n}^{3}}}\] \[=\frac{\Sigma n}{\Sigma {{n}^{3}}}\] \[=\frac{n(n+1)/2}{{{[n(n+1)/2]}^{2}}}\] \[=\frac{2}{n(n+1)}\] \[=2\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[\therefore \] \[{{S}_{n}}={{T}_{1}}+{{T}_{2}}+.....+{{T}_{n}}\] \[=2\left( \frac{1}{1}-\frac{1}{2} \right)+2\left( \frac{1}{2}-\frac{1}{3} \right)\] \[+...+2\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[=2\left( 1-\frac{1}{n+1} \right)\] \[=2-\frac{2}{n+1}\le 2\] \[\left( \because \frac{2}{n+1}\le 1 \right)\]
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