Whoa!!Solve this without a calculator!!-2

There is a super fast way to solve this if you know the identity $\sum_{k=1}^n k^3 = (T_n)^2$ where $T_n$ is the $n$ th triangular number. Substitute $n=10.$ $\sum_{k=1}^{10} k^3 = (T_{10})^2 = 55^2 \\ = \boxed{3025}$

Answer must be 2025 not 3025 as first 10 whole numbers will be 0,1,2,3,4,5,6,7,8,9

We can solve this problem with the help of a generalized formula which says that the sum of the cubes of the first n numbers is

([n(n+1)]/2) ^2

Thus the required answer is [(10X11)/2]^2 =3025