The sum to n terms of the series 3-12-5-12-22-7-12-22-32- is

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Geometric Progression

The sum to ...

Question

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

\frac{6 n}{n + 1}

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\frac{9 n}{n + 1}

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\frac{12 n}{n + 1}

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\frac{3 n}{n + 1}

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Solution

The correct option is A $\frac{6 n}{n + 1}$
The observation is important ...we can see that the numerator general term is $2 n + 1$
while the denominator is the sum of $n^{2}$ terms so the generalized term
$T n = (2 n + 1) / [n (n + 1) (2 n + 1) / 6] = 6 / [n (n + 1)] = 6 [(1 / n) - (1 / (n + 1))]$
Therefore when we add five terms we can see all terms will cancel out except the first and the last term.
So $S = 6 [1 - 1 / (n + 1)] = 6 n / (n + 1)$
Option A

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The sum to n terms of the series 312+512+22+712+22+32+... is

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