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JEE Mains · Maths · STD 11 - 8. sequence and series
Let \({S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + \frac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + ........ + \frac{{1 + 2 + ..... + n}}{{{1^3} + {2^3} + ..... + {n^3}}}\) , If \(100\, S_n\, = n\) , then \(n\) is equal to
- A \(199\)
- B \(99\)
- C \(200\)
- D \(19\)
Answer & Solution
Correct Answer
(A) \(199\)
Step-by-step Solution
Detailed explanation
\({T_n} = \frac{{\frac{{n\left( {n + 1} \right)}}{2}}}{{{{\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)}^2}}}\) \({T_n} = \frac{2}{{n\left( {n + 1} \right)}}\) \({S_n} = \sum {{T_n} = 2\sum\limits_{n = 1}^n {\left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)} } \)…
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