Question

What is the sum of the special series $1^3+2^3+3^3+.........+{40}^3$?

• A. $672400$
• B. $572400$
• C. $472400$
• D. $372400$

Answer 1

The correct option is A $672400$

We know that ${\sum }_{i=1}^n{n}^3={\left[\frac{n(n+1)}{2}\right]}^2$

Here $n$ is the number of terms, so $n=40$

$\therefore S_n={\left[\frac{40\times 41}{2}\right]}^2=672400$