Question 436 of 519MediumMCQ
Let
Sn = \[{1 \over {{1^3}}}\] +13+231+2+13+23+331+2+3+.........+13+23+......+n31+2+.......+n.
Solutions · 1
Official solution
nth term, Tn = \[{{1 + 2 + .... + n} \over {{1^2} + {2^2} + .... + {n^2}}}\]
Tn = \[{{{{n\left( {n + 1} \right)} \over 2}} \over {{{\left( {{{n\left( {n + 1} \right)} \over 2}} \right)}^2}}}\]
\[ \Rightarrow \] Tn = \[{2 \over {n\left( {n + 1} \right)}}\] = \[2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]\]
\[ \therefore \] Sn = \[\sum {{T_n}} \]
= \[2\sum\limits_{n = 1}^n {\left[ {{1 \over n} - {1 \over {n + 1}}} \right]} \]
= \[2\left( {1 - {1 \over n}} \right)\]
= \[{{{2n} \over {n + 1}}}\]
Given that,
100 Sn = n
\[ \Rightarrow \] 100 \[ \times \] \[{{{2n} \over {n + 1}}}\] = n
\[ \Rightarrow \] n + 1 = 200
\[ \Rightarrow \] n = 199
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