The sum to n terms of the series 3 1^ 2 + 5 1^ 2 +2^ 2 + 7 1^ 2 +2^ 2 +3^ 2 + , is
Mathematics · JEE Main · NTA Exams — Progression and Series
The sum to n terms of the series
\(\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+\ldots \ldots,\) is
\(\frac{3}{1^{2}}+\frac{5}{1^{2}+2^{2}}+\frac{7}{1^{2}+2^{2}+3^{2}}+\ldots \ldots,\) is
- \(\frac{6 n}{n+1}\)
- \(\frac{9 \mathrm{n}}{\mathrm{n}+1}\)
- \(\frac{12 n}{n+1}\)
Answer
(A) 6 n n+1
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