Question

If sum of first n terms of the series ${{1}^{2}}+2\cdot {{2}^{2}}+{{3}^{2}}+2\cdot
{{4}^{2}}+{{5}^{2}}+2\cdot {{6}^{2}}+...$ is $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$, when n is even.
When n is odd, the sum is –
(a) $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$
(b) $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
(c) $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$
(d) $\dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{2}$

Answer 1

Hint: We must first assume n = 2m+1, because we need to find the sum for odd values of n. Then we can use the formula for the sum of squares of first n natural numbers, that is, $\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. We can then simplify the series and substitute n to find the correct answer.

Complete step-by-step solution:
Let us first have a thorough look on the given series. We have,
${{1}^{2}}+2\cdot {{2}^{2}}+{{3}^{2}}+2\cdot {{4}^{2}}+{{5}^{2}}+2\cdot {{6}^{2}}+...$
We can also express the same series as
${{1}^{2}}+{{2}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{4}^{2}}+{{5}^{2}}+{{6}^{2}}+{{6}^{2}}+...$
Now, we can regroup these terms as following,
$\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+{{6}^{2}}+... \right)+\left( {{2}^{2}}+{{4}^{2}}+{{6}^{2}}+... \right)$
Here, in this question, we need to calculate the sum of this series up to n terms, when n is odd.
So, let us assume n = 2m + 1.
So, the series we have looks like,
${{1}^{2}}+2\cdot {{2}^{2}}+{{3}^{2}}+2\cdot {{4}^{2}}+{{5}^{2}}+2\cdot {{6}^{2}}+...+2\cdot {{\left( 2m \right)}^{2}}+{{\left( 2m+1 \right)}^{2}}$
Using the above approach, we ca rewrite this series as,
$\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+{{6}^{2}}+...+{{\left( 2m \right)}^{2}}+{{\left( 2m+1 \right)}^{2}} \right)+\left( {{2}^{2}}+{{4}^{2}}+{{6}^{2}}+...+{{\left( 2m \right)}^{2}} \right)$
Let us take 4 as common form the second part of this series. Thus, we get
$\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}}+{{6}^{2}}+...+{{\left( 2m \right)}^{2}}+{{\left( 2m+1 \right)}^{2}} \right)+4\left[ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{m}^{2}} \right]$
We can also represent this as, $\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}$.
We all know very well that the sum of squares of first n natural numbers, that is, $\sum{{{n}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$.
Hence, we can write,
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{\left( 2m+1 \right)\left( 2m+1+1 \right)\left( 2\left( 2m+1 \right)+1 \right)}{6}+\dfrac{4m\left( m+1 \right)\left( 2m+1 \right)}{6}$
On simplification on the right hand side, we can write,
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{\left( 2m+1 \right)\left( 2m+2 \right)\left( 4m+3 \right)}{6}+\dfrac{4m\left( m+1 \right)\left( 2m+1 \right)}{6}$
WE can also write the above equation as
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{2\left( 2m+1 \right)\left( m+1 \right)\left( 4m+3 \right)}{6}+\dfrac{4m\left( m+1 \right)\left( 2m+1 \right)}{6}$
Let us now take $\dfrac{2\left( m+1 \right)\left( 2m+1 \right)}{6}$ as common from the two terms on the right hand side,
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{2\left( m+1 \right)\left( 2m+1 \right)}{6}\left[ 4m+3+2m \right]$
Thus, on simplification, we can easily write
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{2\left( m+1 \right)\left( 2m+1 \right)\left( 6m+3 \right)}{6}$
Hence, we get
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{\left( 2m+2 \right)\left( 2m+1 \right)\left( 2m+1 \right)}{2}$
Now, let us substitute the value n = 2m + 1. Hence, we get
$\sum{{{\left( 2m+1 \right)}^{2}}}+4\sum{{{m}^{2}}}=\dfrac{\left( n+1 \right)\left( n \right)\left( n \right)}{2}$
Hence, the sum is $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$.
So, option (a) is the correct answer.

Note: We can also solve this problem using an objective approach by using the trial and error methodology. We can take examples of small odd integers to verify and find the correct option. WE must also note that we need to find the sum for n terms and not up to the number n, which is the same in this case, but may be different in other similar questions.