The sum to n terms of the infinite series $1 3^{2} + 2 5^{2} + 3 7^{2} + . . . \infty$ is

\frac{n}{6} (n + 1) (6 n^{2} + 14 n + 7)

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\frac{n}{6} (n + 1) (2 n + 1) (3 n + 1)

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4 n^{3} + 4 n^{2} + n

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None of these

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Solution

The correct option is D None of these

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Similar questions

^{n} C_{1} + 2 \cdot 5^{n} C_{2} + 3 \cdot 5^{2}^{n} C_{3} + \dots

n

Sum of $n$ terms of the series whose $n^{t h}$ term is $2^{n - 1} + 8 n^{3} - 6 n^{2}$ .

n

n^{t h}

4 n (n^{2} + 1) - (6 n^{2} + 1)

Find the sum of $1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . . . . n {.1}^{2}$

n^{4} + 6 n^{3} + 5 n^{2}

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