Sum of cubes of first 'n' natural numbers — lesson. Mathematics State Board, Class 10.

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The natural numbers are

$1$ ,

$2$ ,

$3$ ,

$4$ , …

We need to find value of

$1^3 + 2^3 + 3^3 + …. + n^3$ .

Consider the identity

$(x + 1)^{k + 1} - x^{k + 1}$ .

Since the power of all the numbers is

$3$ , put

$k = 3$ in the above identity.

$(x + 1)^{3 + 1} - x^{3 + 1} = (x + 1)^4 - x^4$

$= ((x + 1)^2)^2 - x^4$

$= (x^2 + 2x + 1^2)^2 - x^4$

$= x^4 + 4x^2 + 1^4 + 4x^3 + 4x + 2x^2 - x^4$

$= 4x^3 + 6x^2 + 4x + 1$

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

Now, substitute

$x = 1, 2, 3, … n$ in equation (I).

When

$x = 1$ ,

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

When

$x = 2$ ,

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

When

$x = 3$ ,

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

$\vdots$

When

$x = n - 1$ ,

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

When

$x = n$ ,

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

Add all the above equations of

$x$ values.

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

$2^4 + 3^4 + 4^4 + \cdots + n^4 + (n + 1)^4 - (1^4 + 2^4 + 3^4 + \cdots + (n - 1)^4 + n^4) = 4(1^3 + 2^3 + 3^3 + … + (n - 1)^3 + n^3) + 6(1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) + 4(1 + 2 + 3 + \cdots (n - 1) + n) + (1 + 1 + 1 + ... n \ times )$

By cancelling the same terms with opposite signs on the LHS, we get:

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

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

Apply sum of first

$n$ terms of natural numbers formula in

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

And, apply sum of squares of first

$n$ terms of natural numbers formula in

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

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

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

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

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

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

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

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

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

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

Therefore,

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

Sum of cubes of first

$n$ natural numbers

$=$

${(\frac{n (n + 1)}{2})}^{2}$

Important!

Amicable numbers, or friendly numbers, are a pair of numbers whose sum of proper divisors equals the other.

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5. Sum of cubes of first 'n' natural numbers

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