Q. The sum of cubes of the first ' $n$ ' natural numbers $=$ $(\frac{n ( n + 1 )}{2})^{2} [$ i.e. $1^{3} + 2^{3} + 3^{3} + \dots + n^{3} = [\frac{n ( n + 1 )}{2}]^{2}]$ . Using this result find the value of $1 1^{3} + 1 2^{3} + 1 3^{3} + \dots$ $+ 2 0^{3}$ .

117 141 Squares and Square Roots and Cubes and Cube Roots Report Error

A

45,225

B

41,075

C

38,425

D

52,125

Solution:

Correct answer is (b) 41,075