Sum of cubes! there is a shortcut

$1^3 +2^3 + 3^3 + 4^3 + 5^3 + 6^3+ \ldots +100^3$

$\frac{n^2(n+1)^2}{4}$

Where $n = 100$

$=\frac{100^2(100+1)^2}{4}$

$=\frac{100^2(101)^2}{4}$

$=\frac{102010000}{4}$

$= \boxed{25502500}$

Sum of cubes is [n(n+1)/2]^2 Therefore, [100*101/2]^2 =5050^2 =25502500

Use $1^3+2^3+3^3\cdots n^3=(1+2+3\cdots n)^2$

$1^3+2^3+3^3+...+n^3 =(1+2+3+4+...+n)^2 =\frac{n(n+1)}{2}$ ^2 - Since n=100, the answer is $\frac{100(101)}{2}$ ^2=5050^2=25502500