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ISI2014-DCG-23

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Answer: $\mathbf A$

Explanation:

The above series forms an $\mathbf {AP}$

where, $\mathrm {a = 3, \;d = 8,\;a_n = (8n-5)}$

So, sum of n terms of $\mathbf {AP}$ is given by:

$\begin {align} \mathrm {S_n} &= \mathrm {\frac{n}{2}[a + a_n] }\\&= \mathrm {\frac{n}{2}[3 + 8n-5] }\\&=\mathrm { \frac{n}{2}[8n-2] }\\&= \mathrm {n(4n-1)}\end {align}$

$\therefore \mathrm{S_n = 4n^2-n}$

$\therefore \mathbf A$ is the right option.

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Jeet's answer is good enough, but if you are looking for another approach, it could be : 

3 + 11 + ... + (8n-5) = $\sum (8n-5) = 8n(n+1)/2 - 5n = 4n^2 - n$

The summation function is $n(n+1)/2$ instead of $n(n-1)/2$ because in summation, the starting value of n is taken as 0, whereas, we want it to start from 1 here.

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