Chapter 9.5 - Sum of The Squares of First n Natural Numbers

In the previous section, we completed a discussion on geometric progression and geometric mean. In this section, we will see sum to n terms of special series.

We have to find the sum of three series. They are:

A. 1 + 2 + 3 +. . . + n
   ♦ This is the sum of first n natural numbers.

B. 1² + 2² + 3² +. . . + n²
   ♦ This is the sum of squares of the first n natural numbers.

C. 1³ + 2³ + 3³ +. . . + n³
   ♦ This is the sum of cubes of the first n natural numbers.

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A. 1 + 2 + 3 +. . . + n

This sum can be calculated in 3 steps:
1. The given series is the series related to the sequence 1, 2, 3, . . . , n
2. This sequence is an A.P with a = 1 and d = 1
• So sum to n terms will be given by:
$\begin{array}{ll}
{}&{S_n}
&{}={}& {\frac{n}{2}[2a+(n-1)d]}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {\frac{n}{2}[2 × 1+(n-1) × 1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {\frac{n}{2}[2+n-1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {\frac{n}{2}[n+1]}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

3. We can write it as a formula:
Sum of first n natural numbers = $\frac{n}{2}[n+1]$

---

B. 1² + 2² + 3² +. . . + n²

This sum can be calculated in 8 steps:
1. Consider the identity: (a+b)³ = a³ + 3a²b + 3 ab² + b³
Let us put a = k and b = -1. We get:

$\begin{array}{ll}
{}&{[k+(-1)]^3}
&{}={}& {[k-1]^3}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3 ~+~ 3 × k^2 × -1 ~+~3 × k × (-1)^2~+~(-1)^3}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\


{}&{}
&{}={}& {k^3 ~-~ 3 k^2 ~+~3k~-~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

\end{array}$

2. Subtracting (k-1)³ from k³, we get:

$\begin{array}{ll}
{}&{k^3~-~(k-1)^3}
&{}={}& {k^3~-~\left(k^3 ~-~ 3 k^2 ~+~3k~-~1 \right)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {k^3~-~k^3 ~+~ 3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\

{}&{}
&{}={}& {3 k^2 ~-~3k~+~1}
&{}& {}
&{}{}& {}
&{}& {}&{}& {} &{} \\
\end{array}$

• We can use this as an identity:
$k^3~-~(k-1)^3~=~3 k^2 ~-~3k~+~1$

3. In the above identity, let us put k = 1, 2, 3, . . . , n successively. We get:

$\begin{array}{ll}
{\text{When k = 1,}}&{1^3~-~(1-1)^3}
&{}={}& {1^3~-~0^3}
&{}={}& {3 × 1^2 ~-~3 × 1~+~1}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 2,}}&{2^3~-~(2-1)^3}
&{}={}& {2^3~-~1^3}
&{}={}& {3 × 2^2 ~-~3 × 2~+~1}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 3,}}&{3^3~-~(3-1)^3}
&{}={}& {3^3~-~2^3}
&{}={}& {3 × 3^2 ~-~3 × 3~+~1}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{\text{When k = 4,}}&{4^3~-~(4-1)^3}
&{}={}& {4^3~-~3^3}
&{}={}& {3 × 4^2 ~-~3 × 4~+~1}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{-}&{-}
&{}& {-}
&{}& {-}
&{}& {-}
&{}& {}&{}& {} &{} \\

{\text{When k = n,}}&{n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3 × n^2 ~-~3 × n~+~1}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

4. Picking the first and last items from each line, we get:

$\begin{array}{ll}
{}&{}
&{}& {1^3~-~0^3}
&{}& {}
&{}={}& {3(1)^2~-~3(1)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {2^3~-~1^3}
&{}& {}
&{}={}& {3(2)^2~-~3(2)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {3^3~-~2^3}
&{}& {}
&{}={}& {3(3)^2~-~3(3)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {4^3~-~3^3}
&{}& {}
&{}={}& {3(4)^2~-~3(4)~+~1}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {-}
&{}& {}
&{}& {-}
&{}& {}&{}& {} &{} \\

{}&{}
&{}& {n^3~-~(n-1)^3}
&{}& {}
&{}={}& {3(n)^2~-~3(n)~+~1}
&{}& {}&{}& {} &{} \\

\end{array}$

5. In the above result in (4), let us add all terms on the left side.
• We see that diagonal elements will get cancelled:
   ♦ 1³ will get cancelled by -1³.
   ♦ 2³ will get cancelled by -2³.
   ♦ 3³ will get cancelled by -3³.
   ♦ so on . . .
• So only 0³ and n³ will remain.
• Thus the sum of all terms on the left side is: (n³ - 0³) = n³.

6. In the result in (4), let us add all terms on the right side. We get:
3(1² + 2² + 3² +. . . + n²) - 3(1 + 2 + 3 +. . . + n) + (1+1+1+ . . . n times)
• This can be written in a shortened form using sigma notations:
$3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n$

7. Equating the results in (5) and (6), we get:

$\begin{array}{ll}
{}&{n^3}
&{}={}& {3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~-~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~+~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~3 \sum\limits_{k\,=\,0}^{k\,=\,n}{k}~-~n}
&{\color {green} {\text{- - - - (a)}}}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {n^3~+~\frac{3n(n+1)}{2}~-~n}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 +3n(n+1) - 2n}{2}}
&{}& {}
 \\

{\Rightarrow}&{3 \sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + 3n - 2n}{2}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{2n^3 + 3n^2 + n}{6}}
&{}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n^2 + 3n + 1)}{6}}
&{\color {green} {\text{- - - - (b)}}}& {}
 \\

{}&{}
&{}& {}
&{}& {}
 \\

{\Rightarrow}&{\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}}
&{}={}& {\frac{n(2n + 1) (n+1)}{6}}
&{}& {}
\\

\end{array}$

Remarks:
(i) The line marked as (a):
• $\sum\limits_{k\,=\,0}^{k\,=\,n}{k}$ is in fact the sum of first n natural numbers.
• So our first result A that we saw at the beginning of this section can be used.
(ii) The line marked as (b):
• (2n² + 3n +1) can be written as (2n² + 2n + n + 1)
• But (2n² + 2n + n + 1) = [2n(n+1) + (n+1)] = [(2n+1)(n+1)]
So we get: (2n² + 3n + 1) = (2n+1)(n+1)

Another method:
• Find the solutions of the quadratic equation 2n² + 3n + 1 = 0
• The solutions are: n = $-\frac{1}{2}$ and n = -1
• So we can write: $2n^2 + 3n + 1 ~=~ \left(n+ \frac{1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(\frac{2n+1}{2} \right) (n+1) ~=~0$
$\Rightarrow~\left(2n+1 \right) (n+1) ~=~0$

8. We can write it as a formula:
Sum of squares of the first n natural numbers
= 1² + 2² + 3² +. . . + n²

= $\sum\limits_{k\,=\,1}^{k\,=\,n}{k^2}~=~\frac{n(2n + 1) (n+1)}{6}$

---

In the next section we will see sum of the cubes of first n natural numbers.

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