Sum Of Powers Of Natural Numbers

$\underline{I.\ Introduction}$

We learnt about the sum of the first $n$ natural numbers .

Today we will learn about the sums of various powers of the first $n$ natural numbers.

We will be looking as series like:

$1^2 + 2^2 + 3^2 ...$

$1^3 + 2^3 + 3^3 ...$

and so on.

$\underline{II.\ Sum\ Of\ Squares\ Of\ Natural\ Numbers}$

We will derive the formula for the sum of the squares of the first $n$ natural numbers using the identity:

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

Substituting $n = 1,\ 2,\ 3...n$, we can write:

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

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

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

$.$

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

If we add all the equations above, all terms of the left hand side will cancel out except $0^3$ and $n^3$

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Therefore, we get:

$n^3 - 0^3 = 3.1^2 + 3.2^2 + 3^2 + ... + n^2 - 3.1 - 3.2 - 3.3 + ... - 3n + n$

$\Rightarrow n^3 = 3(1^2 + 2^2 + 3^2 + ... + n^2) - 3(1 + 2 + 3 + ... + n) + n$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n^3 + 3(1 + 2 + 3 + ... + n) - n$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n^3 + 3\dfrac{n(n+1)}{2} - n$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n\left\{n^2 + 3\dfrac{(n+1)}{2} - 1\right\}$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n\left\{\dfrac{2n^2 + 3n + 1}{2}\right\}$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n\left\{\dfrac{2n^2 + 2n + n + 1}{2}\right\}$

$\Rightarrow 3(1^2 + 2^2 + 3^2 + ... + n^2) = n\left\{\dfrac{(n + 1)(2n+1)}{2}\right\}$

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

$\therefore 1^2 + 2^2 + 3^2 + ... + n^2 = \dfrac{n(n + 1)(2n+1)}{6}$

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$\underline{III.\ Sum\ Of\ Cubes\ Of\ Natural\ Numbers}$

Similar to the above approach, to find the sum of the cubes of natural numbers, we can use the identity:

$n^4 - (n- 1)^4 = n^4 - (n^4 - 4n^3 +6n^2 - 4n + 1) = 4n^3 - 6n^2 + 4n - 1$

As before, substituting $n = 1,\ 2,\ 3,\ ... \ n$ we get:

$1^4 - 0^4 = 4.1^3 - 6.1^2 + 4.1 - 1$

$2^4 - 1^4 = 4.2^3 - 6.2^2 + 4.2 - 1$

$.$

$n^4 - (n-1)^4 = 4.n^3 - 6.n^2 + 4.n - 1$

Adding all the equations together, we get:

$n^4 = 4(1^3 + 2^3 + ... + n^3) - 6(1^2 + 2^2 + ... + n^2) + 4(1 + 2 + ... + n) - n$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n^4 + 6(1^2 + 2^2 + ... + n^2) - 4(1 + 2 + ... + n) + n$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n^4 + 6\dfrac{n(n + 1)(2n+1)}{6} - 4\dfrac{n(n+1)}{2} + n$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n\{n^3 + (n + 1)(2n+1) - 2(n+1) + 1\}$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n\{n^3 + (2n^2 + 3n + 1) - 2(n+1) + 1\}$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n\{n^3 + 2n^2 + 3n + 1 - 2n - 2 + 1\}$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n\{n^3 + 2n^2 + n\}$

$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n^2\{n^2 + 2n + 1\}$

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$\Rightarrow 4(1^3 + 2^3 + ... + n^3) = n^2(n + 1)^2$

$\Rightarrow 1^3 + 2^3 + ... + n^3 = \dfrac{n^2(n + 1)^2}{4}$

$\Rightarrow 1^3 + 2^3 + ... + n^3 = \left\{\dfrac{n(n + 1)}{2}\right\}^2$

We can observe from the above that:

$\sum\limits_{i=1}^{n}{i^3} = \left(\sum\limits_{i=1}^{n}{i}\right)^2$

We can also use the above approach to find the formula for the sum of any higher power of the natural numbers, but we have to find them in sequence, that is, to find the sum of the $k\xasuper{th}$ power of the first $n$ natural numbers, we need to know the expressions for the sum of all powers of natural numbers upto $k-1$.

For now we will need to know the sum of the $3\xasuper{rd}$ powers of natural numbers.