Question

# Find the sum of the following series to n terms. The series is:${{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......$

Hint: Use the sum of the series formula for squared terms which is $\sum\limits_{k=1}^{n}{{{\left( k \right)}^{2}}}=\dfrac{k(k+1)(2k+1)}{6}$. Convert the given series in the above sigma notation form.

${{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......$
Here the ${n}^{th}$ term will be ${{(2n)}^{2}}$. So, the series up to n terms will be written as: ${{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......+{{(2n)}^{2}}$
\begin{align} & \Rightarrow {{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......+{{(2n)}^{2}} \\ & \Rightarrow ({{2}^{2}}\times {{1}^{2}})+({{2}^{2}}\times {{2}^{2}})+({{2}^{2}}\times {{3}^{2}})+({{2}^{2}}\times {{4}^{2}})+.......({{2}^{2}}\times {{n}^{2}}) \\ \end{align}
Now taking ${{(2)}^{2}}$ common, we get:
\begin{align} & \Rightarrow ({{2}^{2}}\times {{1}^{2}})+({{2}^{2}}\times {{2}^{2}})+({{2}^{2}}\times {{3}^{2}})+({{2}^{2}}\times {{4}^{2}})+.......({{2}^{2}}\times {{n}^{2}}) \\ & \Rightarrow {{2}^{2}}\times ({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}}) \\ \end{align}
Now, we know that the series of form $({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}$.
\begin{align} & \Rightarrow {{2}^{2}}\times ({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}}) \\ & \Rightarrow {{2}^{2}}\times \dfrac{n(n+1)(2n+1)}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because [({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}] \\ & \Rightarrow \dfrac{2n(n+1)(2n+1)}{3} \\ \end{align}
So finally the sum of the given series ${{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......+{{(2n)}^{2}}$ is $\dfrac{2n(n+1)(2n+1)}{3}$.
Note: We have to be careful in applying the formula for the sum of the series. Also, while splitting the term of the series the ${n}^{th}$ term is ${{(2n)}^{2}}$ and not ${{(n)}^{2}}$.