1. Tardigrade
  2. Question
  3. Mathematics
  4. The sum of cubes of the first ' n ' natural numbers = ((n(n+1)/2))2[. i.e. .13+23+33+ ldots+n3=[(n(n+1)/2)]2]. Using this result find the value of 113+123+133+ ldots +203.

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The sum of cubes of the first ' $n$ ' natural numbers $=$ $\left(\frac{n(n+1)}{2}\right)^2\left[\right.$ i.e. $\left.1^3+2^3+3^3+\ldots+n^3=\left[\frac{n(n+1)}{2}\right]^2\right]$. Using this result find the value of $11^3+12^3+13^3+\ldots$ $+20^3$.

Squares and Square Roots and Cubes and Cube Roots

Solution:

Correct answer is (b) 41,075