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Series

The sum of th...

Question

The sum of the series $3^{2} + 6^{2} + 9^{2} + 12^{2} + . . .$ upto $n$ terms is $S_{n}$ , then $S_{5}$ is equal to

495

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720

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260

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630

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Solution

The correct option is D $495$
$3^{2} + 6^{2} + . . . . \sum (3 n)^{2} 9 \sum n^{2} 9 \frac{n (n + 1) (2 n + 1)}{6} n = 5 \frac{9 \times 5 \times 6 \times 11}{6} = 495$

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Similar questions

1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + 7^{2} + . . .

S_{n} = \{\begin{cases} \frac{n {(n + 1)}^{2}}{2}, if n is even \\ \frac{n^{2} (n + 1)}{2}, if n is odd \end{cases}

1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + 5^{2} + 2 \times 6^{2} + . . . . .

\frac{n (n + 1)^{2}}{2}

n

n

\frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + . . . . . + \frac{1 + 2 + . . . . . + n}{1^{3} + 2^{3} + . . . . . + n^{3}}

n

S_{n}, n = 1, 2, 3, . . . . .

S_{n}

\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \dots

n

S_{n} .

4 S_{\infty}

The sum of the series 3² + 6² + 9² + ... + (3n)² is:

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