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**

**Q1: What is sum of squares?**

Answer: The sum of squares formula is generally referred by the sum of squares of first n natural numbers. It means 1^{2} + 2^{2} + 3^{2} + … + n^{2} = Σ n^{2}. The formula of sum of squares is as follows: 1^{2} + 2^{2} + 3^{2} + … + n^{2} = [n(n+1)(2n+1)] / 6.

**Q2: What is the formula for the sum of squares of even natural numbers?**

Answer: The sum of squares of even natural numbers formula is Σ(2n)^{2} = 2^{2} + 4^{2} + 6^{2} + … + (2n)^{2} = [2n(n + 1)(2n + 1)] / 3.

**Q3: Write down the sum of squares of odd natural numbers formula.**

Answer: The formula of the sum of squares of odd natural numbers is Σ(2n-1)^{2} = 1^{2} + 3^{2} + 5^{2} + … + (2n-1)^{2} = [n(2n+1)(2n-1)] / 3.