Sum of squares of natural numbers

In this section, we will discuss the formulas of the sum of the squares of natural numbers. These formulas are very useful in various competitive exams.

Sum of squares of first n natural numbers:

The sum of squares of consecutive natural numbers is determined by the formula below:

Prove that: $1^2+2^2+3^2+\cdots+n^2$ $=\dfrac{n(n+1)(2n+1)}{6}$

Proof: Let $S$ denote the desired sum. That is, $S=1^2+2^2+\cdots+n^2.$ We will use the following fact:

$n^3-(n-1)^3=3n^2-3n+1$ $\cdots$ (I)

This is obtained by applying the formula $a^3-b^3$$=(a-b)(a^2+ab+b^2)$

Using (I) we get the following relations:

$1^3-0^3=3.1^2-3.1+1$

$2^3-1^3=3.2^2-3.2+1$

$3^3-2^3=3.2^2-3.3+1$

$\quad \vdots \quad \quad \quad \vdots$

$n^3-(n-1)^3=3n^2-3n+1$

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Adding, we get that

$n^3-0^3$ $=3(1^2+2^2+\cdots +n^2)$ $-3(1+2+\cdots +n)+$ $(1+1+\cdots$ till n terms)

$\Rightarrow n^3=3S-3 \frac{n(n+1)}{2}+n$, since the sum of the first n natural numbers is n(n+1)/2.

$\Rightarrow 3S=3 \frac{n(n+1)}{2}+n^3-n$

$\Rightarrow 3S=3 \frac{n(n+1)}{2}$ $+n(n-1)(n+1)$

$\Rightarrow 3S=n(n+1)[\frac{3}{2}$ $+(n-1)]$

$\Rightarrow 3S=n(n+1)[\frac{3+2n-2}{2}]$

$\Rightarrow 3S=n(n+1)[\frac{2n+1}{2}]$

$\Rightarrow S=\frac{n(n+1)(2n+1)}{6}$

Remark: Note that the above sum $1^2+2^2+\cdots +n^2$ can be written as $\sum_{k=1}^n k^2$

SOLVED EXAMPLES

Problem 1: Find the sum of the squares of the first 100 natural numbers.

Solution:

We need to find the sum $1^2+2^2+\cdots+100^2$

By the above formula, we know that the sum of the squares of first n natural numbers is $\frac{n(n+1)(2n+1)}{6}$

So we have the sum =

$1^2+2^2+\cdots+100^2$ $=\frac{n(n+1)(2n+1)}{6}$ where $n=100$

$=\frac{100(100+1)(2.100+1)}{6}$

$=\frac{100 \times 101 \times 201}{6}$

$=50 \times 101 \times 67$

$=338350$

Sum of squares of first n even natural numbers:

Prove that: $2^2+4^2+6^2+\cdots+(2n)^2$ $=\frac{2n(n+1)(2n+1)}{3}$

Proof: Note that

$2^2+4^2+6^2+\cdots+(2n)^2$

$=\sum_{k=1}^n(2k)^2$

$=\sum_{k=1}^n4k^2$

$=4\sum_{k=1}^nk^2$

$=4 \times \dfrac{n(n+1)(2n+1)}{6}$ as the sum of squares of first n natural numbers is n(n+1)(2n+1)/6.

$=\dfrac{2n(n+1)(2n+1)}{3}$

Sum of squares of first n odd natural numbers:

Prove that: $1^2+3^2+5^2+\cdots+(2n-1)^2$ $=\dfrac{n(2n-1)(2n+1)}{3}$

FAQs on Sum of Squares