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:
$1^3-0^3=3.1^2-3.1+1$
$2^3-1^3=3.2^2-3.2+1$
$3^3-2^3=3.2^2-3.3+1$
$\quad \vdots \quad \quad \quad \vdots$
$n^3-(n-1)^3=3n^2-3n+1$
——————————————-
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
Answer: The sum of squares formula is generally referred by the sum of squares of first n natural numbers. It means 12 + 22 + 32 + … + n2 = Σ n2. The formula of sum of squares is as follows: 12 + 22 + 32 + … + n2 = [n(n+1)(2n+1)] / 6.
Answer: The sum of squares of even natural numbers formula is Σ(2n)2 = 22 + 42 + 62 + … + (2n)2 = [2n(n + 1)(2n + 1)] / 3.
Answer: The formula of the sum of squares of odd natural numbers is Σ(2n-1)2 = 12 + 32 + 52 + … + (2n-1)2 = [n(2n+1)(2n-1)] / 3.