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 cubes of first n natural numbers:
We determine the sum of cubes of consecutive natural numbers by the following formula:
| Prove that: $1^3+2^3+3^3+\cdots+n^3$ $=[\frac{n(n+1)}{2}]^2$ |
Proof: Let $S$ denote the desired sum. So we have $S=1^3+2^3+\cdots+n^3.$ To prove the formula, we will use the fact below:
$n^4-(n-1)^4=4n^3-6n^2+4n-1$ $\cdots$ (I)
Putting $n=1, 2, \cdots$ we obtain the following relations:
$1^4-0^4=4.1^3-6.1^2+4.1-1$
$2^4-1^4=4.2^3-6.2^2+4.2-1$
$3^4-2^4=4.3^3-6.3^2+4.3-1$
$\quad \vdots \quad \quad \quad \vdots$
$\small{n^4-(n-1)^4=4n^3-6n^2+4n-1}$
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termwise addition implies that
$n^4-0^4$ $=4(1^3+2^3+\cdots n^3)-$ $6(1^2+2^2+\cdots +n^2)+$ $4(1+2+\cdots +n)-$ $(1+1+\cdots$ till n terms)
$\Rightarrow n^3=4S-6 \frac{n(n+1)(2n+1)}{6}+$ $4\frac{n(n+1)}{2}-n$, since the sum the first n natural numbers is n(n+1)/2, the sum of the squares of the first n natural numbers is n(n+1)(2n+1)/6.
$\Rightarrow 4S=n^4+n(n+1)(2n+1)$ $-2n(n+1)+n$
$\Rightarrow 4S= n^4+n(n+1)(2n+1)$ $-2n(n+1)+n$
$\Rightarrow 4S= n^4+n[(n+1)(2n+1)$ $-2(n+1)+1]$
$\Rightarrow 4S= n^4+n[2n^2+3n+1$ $-2n-2+1]$
$\Rightarrow 4S= n^4+n[2n^2+n]$
$\Rightarrow 4S= n^4+2n^3+n^2]$
$\Rightarrow 4S= n^2(n^2+2n+1)$
$\Rightarrow 4S= n^2(n+1)^2$
$\Rightarrow S= \frac{n^2(n+1)^2}{4}$
$\therefore S=[\frac{n(n+1)}{2}]^2$
SOLVED EXAMPLES
Problem 1: Find the sum of the cubes of the first 100 natural numbers.
Solution:
We need to find the sum $1^3+2^3+\cdots+100^3$
By the above formula, we have
$1^3+2^3+\cdots+100^3$ $=[\frac{n(n+1)}{2}]^2$ where $n=100$
So the sum $=[\frac{100(100+1)}{2}]^2$
$=(50 \times 101)^2$
$=5050^2$
