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Question: For a natural number \(n\in N\), the value of \(\sum\limits_{m=1}^{n}{{{m}^{2}}}\) is given by A. ...

For a natural number n∈Nn\in N, the value of ∑m=1nm2\sum\limits_{m=1}^{n}{{{m}^{2}}} is given by
A. m(m+1)(2m+1)6\dfrac{m\left( m+1 \right)\left( 2m+1 \right)}{6}
B. m(m−1)(2m−1)6\dfrac{m\left( m-1 \right)\left( 2m-1 \right)}{6}
C. n(n+1)(2n+1)6\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}
D. None of these.

Explanation

Solution

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 ∑m=1nm2=12+22+32.........n2\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 ∑m=1nm2=12+22+32.........n2=n(n+1)(2n+1)6\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 (n+1)3−n3{{\left( n+1 \right)}^{3}}-{{n}^{3}} and write its value using the formula a3−b3=(a−b)(a2+ab+b2){{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 ∑m=1nm=n(n+1)2\sum\limits_{m=1}^{n}{m}=\dfrac{n\left( n+1 \right)}{2} and simplifying, we get the required value of ∑m=1nm2\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 ∑m=1nm2\sum\limits_{m=1}^{n}{{{m}^{2}}}. To get the answer we should consider (n+1)3−n3{{\left( n+1 \right)}^{3}}-{{n}^{3}}.
Let us consider the term (n+1)3−n3{{\left( n+1 \right)}^{3}}-{{n}^{3}}.
We know the formula a3−b3=(a−b)(a2+ab+b2){{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right). Using this formula with
a=n+1,b=na=n+1,b=n, we get
(n+1)3−n3=(n+1−n)((n+1)2+(n+1)n+n2) (n+1)3−n3=(n2+2n+1+n2+n+n2) (n+1)3−n3=3n2+3n+1→(1) \begin{aligned} & {{\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{aligned}
By substituting n=n−1n=n-1 in the above equation, we get
(n−1+1)3−(n−1)3=3(n−1)2+3(n−1)+1 (n)3−(n−1)3=3(n−1)2+3(n−1)+1→(2) \begin{aligned} & {{\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{aligned}
Likewise, substituting the numbers n=n−2,.....2,1n=n-2,.....2,1 in the equation-1, we get
(n−2+1)3−(n−2)3=3(n−2)2+3(n−2)+1 (n−1)3−(n−2)3=3(n−2)2+3(n−2)+1→(3) \begin{aligned} & {{\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{aligned}
(2+1)3−(2)3=3(2)2+3(2)+1→(4){{\left( 2+1 \right)}^{3}}-{{\left( 2 \right)}^{3}}=3{{\left( 2 \right)}^{2}}+3\left( 2 \right)+1\to \left( 4 \right)
(2)3−(1)3=3(1)2+3(1)+1→(5){{\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
(n+1)3−n3=3n2+3n+1→(1) (n)3−(n−1)3=3(n−1)2+3(n−1)+1→(2) (n−1)3−(n−2)3=3(n−2)2+3(n−2)+1→(3) . . . (2+1)3−(2)3=3(2)2+3(2)+1→(4) (2)3−(1)3=3(1)2+3(1)+1→(5) \begin{aligned} & {{\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{aligned}
Adding all the equations, we can infer that the terms in L.H.S will be cancelled and only (n+1)3−13{{\left( n+1 \right)}^{3}}-{{1}^{3}} is left in the L.H.S. The terms in the R.H.S will become
(n+1)3−13=3(n2+(n−1)2+(n−2)2+......22+12)+3(n+(n−1)+(n−2)....2+1)+(1+1+1.....n terms) (n+1)3−1=3∑m=1nm2+3∑m=1nm+n \begin{aligned} & {{\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{aligned}
We know the formula for the sum of first n natural numbers as ∑m=1nm=n(n+1)2\sum\limits_{m=1}^{n}{m}=\dfrac{n\left( n+1 \right)}{2}. Using this result in the above equation, we get
(n+1)3−1=3∑m=1nm2+3n(n+1)2+n{{\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

& {{\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{aligned}$$ We got relation as $$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}$$.