Question

If the sum of first n terms of an A.P is $c{n^2}$, then the sum of square of these n terms is

A. $\dfrac{{n\left( {4{n^2} - 1} \right){c^2}}}{6}$

B. $\dfrac{{n\left( {4{n^2} + 1} \right){c^2}}}{3}$

C. $\dfrac{{n\left( {4{n^2} - 1} \right){c^2}}}{3}$

D. $\dfrac{{n\left( {4{n^2} + 1} \right){c^2}}}{6}$

Answer 1

In order to solve the above question we must first know what we actually mean by arithmetic series. Suppose a sequence of numbers is arithmetic (that is, it increases or decreases by a constant amount each term), and you want to find the sum of the first n terms. Then we have to apply the formula of sum and get the total number of terms with the help of sum and square it to get the right answer. Doing this will give the right answer.

Complete step-by-step answer:

Denote this partial sum by ${S_n}$. Then we know the general formula of sum of n terms and n-1 terms can be,

${S_n} = \dfrac{n}{2}\left( {a + l} \right)$ and ${S_{n - 1}} = \dfrac{{n - 1}}{2}\left( {a + l} \right)$, where n is the number of terms, a is the first term and l is the last term. The sum of the first n terms of an arithmetic sequence is called an arithmetic series

${t_n} =$total number of n terms

${S_n} =$sum of the n terms

$n =$number of terms

{t_n} = {S_n} - {S_{n - 1}} = c\left\\{ {{n^2} - {{\left( {n - 1} \right)}^2}} \right\\} = c\left( {2n - 1} \right)

On squaring the terms both sides we get,

$\Rightarrow {t_n}^2 = {c^2}\left( {4{n^2} - 4n + 1} \right)$

Then taking the summation of both sides and using the formula of sum of ${n^2},n$ and sum of 1 $n$ times.

We get the equations as,

\Rightarrow \sum\limits_{n = 1}^n {{t^2}_n = } {c^2}\left\\{ {\dfrac{{4n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \dfrac{{4n\left( {n + 1} \right)}}{2} + n} \right\\}

We know that $\sum\limits_{n = 0}^{n = n} {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} ,\sum\limits_{n = 0}^{n = n} n = \dfrac{{n\left( {n + 1} \right)}}{2}$ and $\sum\limits_0^n {1 = n}$

Then on solving further we get the sum of squares of all the terms as,

\Rightarrow \sum\limits_{n = 1}^n {{t^2}_n = } \dfrac{{{c^2}n}}{6}\left\\{ {4\left( {n + 1} \right)\left( {2n + 1} \right) - 12\left( {n + 1} \right) + 6} \right\\}

\Rightarrow \sum\limits_{n = 1}^n {{t^2}_n = } \dfrac{{{c^2}n}}{3}\left\\{ {4{n^2} + 6n + 2 - 6n - 6 + 3} \right\\}

$\Rightarrow \sum\limits_{n = 1}^n {{t^2}_n = } \dfrac{{{c^2}n}}{3}\left( {4{n^2} - 1} \right)$

Therefore the correct answer is $\dfrac{{{c^2}n\left( {4{n^2} - 1} \right)}}{3}$.

So, the correct option is C.

Note : While solving the above question keep in mind that the sum formula for arithmetic series is used .Keep in mind that such questions are easy to solve but calculations and formulas must be clear. We know that $\sum\limits_{n = 0}^{n = n} {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} ,\sum\limits_{n = 0}^{n = n} n = \dfrac{{n\left( {n + 1} \right)}}{2}$ and $\sum\limits_0^n {1 = n}$. Knowing this will help us in most of the problems and will give you the right answer.