Question

The sum of ${{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}$ is

Answer 1

Hint: To solve these types of questions we need to add and subtract the terms before the given series. In this question we will add and subtract ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}}$to the given series ${{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}$. After this, we will use formula ${{1}^{2}}+{{2}^{2}}+...+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ in order to solve it further.

Complete step by step solution:
We are given a series in the form of ${{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}$. To solve this question we will form a series which starts from 1. To convert the given series into such form, we need to add and subtract the terms ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}}$ to our given series. The new series that we got here is in the following form.
$\begin{align}
  & {{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}=\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}} \right)+{{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}-\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}} \right) \\
 & \Rightarrow {{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}}+{{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}-{{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}}\,...(i) \\
\end{align}$
Now, if we watch carefully, we get to know that we have a series as ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{20}^{2}}$ starting from 1 and ending with 20. As we know that for such type of series we can use formula ${{1}^{2}}+{{2}^{2}}+...+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. By taking n = 20 we get
$\begin{align}
  & {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{20}^{2}}=\dfrac{20\left( 20+1 \right)\left( 2\left( 20 \right)+1 \right)}{6} \\
 & \Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{20}^{2}}=\dfrac{20\times 21\times 41}{6} \\
 & \Rightarrow {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....+{{20}^{2}}=2870\,\,....(ii) \\
\end{align}$
Similarly, for series ${{1}^{2}}+{{2}^{2}}+...+{{9}^{2}}=\dfrac{9\left( 9+1 \right)\left( 2\left( 9 \right)+1 \right)}{6}$. Thus, we now have${{1}^{2}}+{{2}^{2}}+...+{{9}^{2}}=\dfrac{9\times 10\times 19}{6}=285$….(iii).
Now, at this point we are going to substitute (ii) and (iii) in (i) so that we get,
${{10}^{2}}+{{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}=2870-285=2585$
Hence, the required value of the given series is 2585.

Note: In case this series was written in the form $\sum\limits_{n=10}^{20}{{{n}^{2}}}$ then also we will solve it as solved above. This is because the series $\sum\limits_{n=10}^{20}{{{n}^{2}}}$and the given series are same. Since, the given series is clearly in squared form this is why we have used the formula ${{1}^{2}}+{{2}^{2}}+...+{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. In case the series started as ${{11}^{2}}+{{12}^{2}}+...+{{20}^{2}}$ then we would have added and subtracted ${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{9}^{2}}+{{10}^{2}}$to it and solved it further. In order to get the right answer of these types of questions we need to solve it with full concentration. Otherwise, we can calculate wrong squares and get towards the wrong answer.