Question

answer this question please by short method.

Answer 1

We know that sum of cubes of first n natural numbers = n^2(n + 1)^2/(4) --- (1)

Now,

Given 1^3 + 2^3 + 3^3+ ..... + 7^3.

Substitute n = 7 in (1), we get

= > 7^2(7 + 1)^2/(4)

= > 49(64)/(4)

= > 3136/4

= > 784.


Therefore 1^3 + 2^3 + 3^3..... + 7^3 = 784.


Hope this helps!