Let 5= 1 n+1 + 1 2(n+1)^ 2 + 1 3(n+1)^ 3 + and T= 1 n - 1 2 n^ 2 + 1 3 n^ 3 - , then :
Mathematics · JEE Main · NTA Exams — Progression and Series
Let \(5=\frac{1}{n+1}+\frac{1}{2(n+1)^{2}}+\frac{1}{3(n+1)^{3}}+\ldots\) and
\(T=\frac{1}{n}-\frac{1}{2 n^{2}}+\frac{1}{3 n^{3}}-\ldots\), then :
- \(5=2 T\)
- \(25=T\)
- \(5=T\)
- \(5=3 T\)
Answer
(A) 5=2 T
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