Question

The cubes of the natural number are grouped as let denote the sum of the cubes in the nth group, then is:

Answer 1

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Let us assume the required sum = S

Therefore, S = 13 + 23 + 33 + 43 + 53 + ................... + n3

Now, we will use the below identity to find the value of S:

n4 - (n - 1)4 = 4n3 - 6n2 + 4n - 1

Substituting, n = 1, 2, 3, 4, 5, ............., n in the above identity, we get

14 - 04 = 4 ∙ 13 - 6 ∙ 12 + 4 ∙ 1 - 1

24 - 14 = 4 ∙ 23 - 6 ∙ 22 + 4 ∙ 2 - 1

34 - 24 = 4 ∙ 33 - 6 ∙ 32 + 4 ∙ 3 - 1

44 - 34 = 4 ∙ 43 - 6 ∙ 42 + 4 ∙ 4 - 1

........ .................... ...............

n4 - (n - 1)4 = 4 . n3 - 6 ∙ n2 + 4 ∙ n - 1

Adding we get, n4 - 04 = 4(13 + 23 + 33 + 43 + ........... + n3) - 6(12 + 22 + 32 + 42 + ........ + n2) + 4(1 + 2 + 3 + 4 + ........ + n) - (1 + 1 + 1 + 1 + ......... n times)

⇒ n4 = 4S - 6 ∙ \(\frac{n(n + 1)(2n + 1)}{6}\) + 4 ∙

n(n+1)

2

- n

⇒ 4S = n4 + n(n + 1)(2n + 1) - 2n(n + 1) + n

⇒ 4S = n4 + n(2n2 + 3n + 1) – 2n2 - 2n + n

⇒ 4S = n4 + 2n3 + 3n2 + n - 2n2 - 2n + n

⇒ 4S = n4 + 2n3 + n2

⇒ 4S = n2(n2 + 2n + 1)

⇒ 4S = n2(n + 1)2

Therefore, S =

n2(n+1)2

4

= {

n(n+1)

2

}2 = (Sum of the first n natural numbers)2

i.e., 13 + 23 + 33 + 43 + 53 + ................... + n3 = {

n(n+1)

2

}2

Thus, the sum of the cubes of first n natural numbers = {

n(n+1)

2

}2