Question

For a natural number $n\in N$, the value of $\sum\limits_{m=1}^{n}{{{m}^{2}}}$ is given by
A. $\dfrac{m\left( m+1 \right)\left( 2m+1 \right)}{6}$
B. $\dfrac{m\left( m-1 \right)\left( 2m-1 \right)}{6}$
C. $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
D. None of these.

Answer 1

Hint: To solve this problem, we should know the formula related to sum of squares of first n natural numbers. The sigma notation given in the question can be written as $\sum\limits_{m=1}^{n}{{{m}^{2}}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}.........{{n}^{2}}$ which is the sum of squares of first n natural numbers. The sum of the first n natural numbers is given by the formula $\sum\limits_{m=1}^{n}{{{m}^{2}}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}.........{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$. To get this formula, we should consider ${{\left( n+1 \right)}^{3}}-{{n}^{3}}$ and write its value using the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. After that, we should replace n by the decreasing numbers until 2. This will give a telescopic series in which terms will be cancelled when we add them. By adding, we will get sum of first n natural numbers and sum of squares of first n natural numbers in the R.H.S of the equation. By using the formula for sum of first n natural numbers as $\sum\limits_{m=1}^{n}{m}=\dfrac{n\left( n+1 \right)}{2}$ and simplifying, we get the required value of $\sum\limits_{m=1}^{n}{{{m}^{2}}}$. The proof related to this formula is done in the below solution part.

Complete step by step answer:
We are asked to find the value of $\sum\limits_{m=1}^{n}{{{m}^{2}}}$. To get the answer we should consider ${{\left( n+1 \right)}^{3}}-{{n}^{3}}$.
Let us consider the term ${{\left( n+1 \right)}^{3}}-{{n}^{3}}$.
We know the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. Using this formula with
$a=n+1,b=n$, we get
$\begin{align}
  & {{\left( n+1 \right)}^{3}}-{{n}^{3}}=\left( n+1-n \right)\left( {{\left( n+1 \right)}^{2}}+\left( n+1 \right)n+{{n}^{2}} \right) \\
 & {{\left( n+1 \right)}^{3}}-{{n}^{3}}=\left( {{n}^{2}}+2n+1+{{n}^{2}}+n+{{n}^{2}} \right) \\
 & {{\left( n+1 \right)}^{3}}-{{n}^{3}}=3{{n}^{2}}+3n+1\to \left( 1 \right) \\
\end{align}$
By substituting $n=n-1$ in the above equation, we get
$\begin{align}
  & {{\left( n-1+1 \right)}^{3}}-{{\left( n-1 \right)}^{3}}=3{{\left( n-1 \right)}^{2}}+3\left( n-1 \right)+1 \\
 & {{\left( n \right)}^{3}}-{{\left( n-1 \right)}^{3}}=3{{\left( n-1 \right)}^{2}}+3\left( n-1 \right)+1\to \left( 2 \right) \\
\end{align}$
Likewise, substituting the numbers $n=n-2,.....2,1$ in the equation-1, we get
$\begin{align}
  & {{\left( n-2+1 \right)}^{3}}-{{\left( n-2 \right)}^{3}}=3{{\left( n-2 \right)}^{2}}+3\left( n-2 \right)+1 \\
 & {{\left( n-1 \right)}^{3}}-{{\left( n-2 \right)}^{3}}=3{{\left( n-2 \right)}^{2}}+3\left( n-2 \right)+1\to \left( 3 \right) \\
\end{align}$
${{\left( 2+1 \right)}^{3}}-{{\left( 2 \right)}^{3}}=3{{\left( 2 \right)}^{2}}+3\left( 2 \right)+1\to \left( 4 \right)$
${{\left( 2 \right)}^{3}}-{{\left( 1 \right)}^{3}}=3{{\left( 1 \right)}^{2}}+3\left( 1 \right)+1\to \left( 5 \right)$
Writing all the equations, we get
$\begin{align}
  & {{\left( n+1 \right)}^{3}}-{{n}^{3}}=3{{n}^{2}}+3n+1\to \left( 1 \right) \\
 & {{\left( n \right)}^{3}}-{{\left( n-1 \right)}^{3}}=3{{\left( n-1 \right)}^{2}}+3\left( n-1 \right)+1\to \left( 2 \right) \\
 & {{\left( n-1 \right)}^{3}}-{{\left( n-2 \right)}^{3}}=3{{\left( n-2 \right)}^{2}}+3\left( n-2 \right)+1\to \left( 3 \right) \\
 & . \\
 & . \\
 & . \\
 & {{\left( 2+1 \right)}^{3}}-{{\left( 2 \right)}^{3}}=3{{\left( 2 \right)}^{2}}+3\left( 2 \right)+1\to \left( 4 \right) \\
 & {{\left( 2 \right)}^{3}}-{{\left( 1 \right)}^{3}}=3{{\left( 1 \right)}^{2}}+3\left( 1 \right)+1\to \left( 5 \right) \\
\end{align}$
Adding all the equations, we can infer that the terms in L.H.S will be cancelled and only ${{\left( n+1 \right)}^{3}}-{{1}^{3}}$ is left in the L.H.S. The terms in the R.H.S will become
$\begin{align}
  & {{\left( n+1 \right)}^{3}}-{{1}^{3}}=3\left( {{n}^{2}}+{{\left( n-1 \right)}^{2}}+{{\left( n-2 \right)}^{2}}+{{......2}^{2}}+{{1}^{2}} \right)+3\left( n+\left( n-1 \right)+\left( n-2 \right)....2+1 \right)+\left( 1+1+1.....\text{n terms} \right) \\
 & {{\left( n+1 \right)}^{3}}-1=3\sum\limits_{m=1}^{n}{{{m}^{2}}+3\sum\limits_{m=1}^{n}{m}}+n \\
\end{align}$
We know the formula for the sum of first n natural numbers as $\sum\limits_{m=1}^{n}{m}=\dfrac{n\left( n+1 \right)}{2}$. Using this result in the above equation, we get
${{\left( n+1 \right)}^{3}}-1=3\sum\limits_{m=1}^{n}{{{m}^{2}}+3\dfrac{n\left( n+1 \right)}{2}}+n$
By simplifying, we get
\[\begin{align}
  & {{\left( n+1 \right)}^{3}}-1=3\sum\limits_{m=1}^{n}{{{m}^{2}}+3\dfrac{n\left( n+1 \right)}{2}}+n \\
 & 3\sum\limits_{m=1}^{n}{{{m}^{2}}={{\left( n+1 \right)}^{3}}-}3\dfrac{n\left( n+1 \right)}{2}-\left( n+1 \right) \\
 & 3\sum\limits_{m=1}^{n}{{{m}^{2}}=\left( n+1 \right)\left( {{\left( n+1 \right)}^{2}}-3\dfrac{n}{2}-1 \right)}=\left( n+1 \right)\left( \dfrac{2{{\left( n+1 \right)}^{2}}-3n-2}{2} \right) \\
 & 3\sum\limits_{m=1}^{n}{{{m}^{2}}=}\dfrac{\left( n+1 \right)}{2}\left( 2{{n}^{2}}+4n+2-3n-2 \right)=\dfrac{\left( n+1 \right)}{2}\left( 2{{n}^{2}}+n \right)=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{2} \\
\end{align}\]
We got relation as
\[3\sum\limits_{m=1}^{n}{{{m}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{2}\]
Hence we can write the sum of the squares of first n natural numbers as
\[\sum\limits_{m=1}^{n}{{{m}^{2}}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]
$\therefore $ We got the required formula for the term \[\sum\limits_{m=1}^{n}{{{m}^{2}}}\] as \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].

So, the correct answer is “Option C”.

Note: Students can make a mistake by choosing the answer as option-A. This is a common error because of the anxiety of knowing the answer. That is why a probable wrong answer is placed in option-A. To avoid this error, we should expand the sigma notation and see whether there are m terms of n terms in the expansion. This is an important formula and students are recommended to remember the formula for the sum of first n natural numbers as \[\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\].