Sum of cubes 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 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 $=[\frac{100(100+1)}{2}]^2$

$=(50 \times 101)^2$

$=5050^2$