Question

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

Answer 1

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.

Complete step-by-step answer:
In the question, it is asked to find the sum of the below series to n terms. The series is:
\[{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......\]
Here the ${n}^{th}$ term will be \[{{(2n)}^{2}}\]. So, the series up to n terms will be written as: \[{{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......+{{(2n)}^{2}}\]
Now, this series can be rewritten, after splitting as follows:
\[\begin{align}
  & \Rightarrow {{2}^{2}}+{{4}^{2}}+{{6}^{2}}+{{8}^{2}}+......+{{(2n)}^{2}} \\
 & \Rightarrow ({{2}^{2}}\times {{1}^{2}})+({{2}^{2}}\times {{2}^{2}})+({{2}^{2}}\times {{3}^{2}})+({{2}^{2}}\times {{4}^{2}})+.......({{2}^{2}}\times {{n}^{2}}) \\
\end{align}\]
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 \[({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}\].
So using the above formula we have:
\[\begin{align}
  & \Rightarrow {{2}^{2}}\times ({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}}) \\
 & \Rightarrow {{2}^{2}}\times \dfrac{n(n+1)(2n+1)}{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because [({{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+........{{n}^{2}})=\dfrac{n(n+1)(2n+1)}{6}] \\
 & \Rightarrow \dfrac{2n(n+1)(2n+1)}{3} \\
\end{align}\]
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 \[{{(n)}^{2}}\].