Question

# Find the sum up to n terms for the following series: ${{3}^{2}}+{{8}^{2}}+{{13}^{2}}+...$(a) $\dfrac{n\left( 50{{n}^{2}}+15n-11 \right)}{6}$(b) $\dfrac{\left( 50{{n}^{2}}+15n-11 \right)}{6}$(c) $\dfrac{n\left( 50{{n}^{2}}+15n+11 \right)}{6}$(d) $\dfrac{n\left( 50{{n}^{2}}-15n-11 \right)}{6}$

Hint: Observe the pattern in the elements of the series. The ith element in this series will be ${{\left( 5i-2 \right)}^{2}}$. Find the summation of this general term from 1 to n. Use the following formulae: $\sum\limits_{i=1}^{n}{{{i}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$, $\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$, and $\sum\limits_{i=1}^{n}{c}=cn$. Simplify the resulting expression to arrive at the final answer.

In this question, we are given the series ${{3}^{2}}+{{8}^{2}}+{{13}^{2}}+...$
We need to find the sum of this series up to n terms.
Let us try to find a pattern in the elements of this series.
We can see that the difference between the roots of the consecutive elements is constant and is equal to 5.
i.e. 8 – 3 = 13 – 8 = 5
Using this, we will try to find a general expression to denote each term in the series.
We can denote the root of the ith element in this series by (5i – 2).
So, the ith element in this series will be ${{\left( 5i-2 \right)}^{2}}$, where i ranges from 1 to n.
So, we need to find $\sum\limits_{i=1}^{n}{{{\left( 5i-2 \right)}^{2}}}$
To find this, let us first expand the expression within the summation.

Expanding the expression within the summation, we will get the following:
$\sum\limits_{i=1}^{n}{\left( 25{{i}^{2}}-20i+4 \right)}$
We will now individually evaluate the summations of the three terms int the above expression.
$\sum\limits_{i=1}^{n}{25{{i}^{2}}}-\sum\limits_{i=1}^{n}{20i}+\sum\limits_{i=1}^{n}{4}$
Now, we know the property for summation of squares of natural numbers:
$\sum\limits_{i=1}^{n}{{{i}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
Also, we know the property of summation of natural numbers:
$\sum\limits_{i=1}^{n}{i}=\dfrac{n\left( n+1 \right)}{2}$
Now, we also know the property for summation of a constant term:
$\sum\limits_{i=1}^{n}{c}=cn$
Using the above properties, we will get the following:
$\sum\limits_{i=1}^{n}{{{\left( 5i-2 \right)}^{2}}}=\dfrac{n\left[ 50{{n}^{2}}+15n-11 \right]}{6}$
Now, we will simplify this expression to arrive at the final answer.
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{60n\left( n+1 \right)}{6}+\dfrac{24n}{6}$
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)-60n\left( n+1 \right)+24n}{6}$
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{n\left( n+1 \right)\left[ 50n+25-60 \right]+24n}{6}$
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{n\left( n+1 \right)\left[ 50n-35 \right]+24n}{6}$
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{n\left[ 50{{n}^{2}}+15n-35+24 \right]}{6}$
$\dfrac{25n\left( n+1 \right)\left( 2n+1 \right)}{6}-\dfrac{20n\left( n+1 \right)}{2}+4n=\dfrac{n\left[ 50{{n}^{2}}+15n-11 \right]}{6}$
So, $\sum\limits_{i=1}^{n}{{{\left( 5i-2 \right)}^{2}}}=\dfrac{n\left[ 50{{n}^{2}}+15n-11 \right]}{6}$
The sum up to n terms for the series: ${{3}^{2}}+{{8}^{2}}+{{13}^{2}}+...$ is $\dfrac{n\left[ 50{{n}^{2}}+15n-11 \right]}{6}$
Hence, option (a) is correct.

Note: In this question, it is very important to identify the pattern in the elements of the series and then find the general terms. Without this, the question will become very lengthy and difficult. In such questions, always look for such patterns and then use the general summation formula to find the answer.