Math Series #1

$1^3 + 2^3 = 1 + 8 = 9 = 3^2$

$1^3 + 2^3 +3^3 = 9 + 27 = 36 = 6^2$

Guess $1^3 + 2^3 + ....+ n^3 = (1+2+...+n)^2$

$n = 1, 2, 3$ are true

Assume $n = k$ is true

$1^3 + 2^3 + ....+ k^3 = (1+2+...+k)^2 = (\cfrac{k(k+1)}{2})^2$

When $n=k+1$ ,

$1^3 + 2^3 + ....+ k^3 + (k+1)^3 \\ = (\cfrac{k(k+1)}{2})^2 + (k+1)^3 \\ = (\cfrac{(k+1)^2(k^2+4(k+1))}{4})\\ = (\cfrac{(k+1)^2(k^2+4k+4))}{4})\\ = (\cfrac{(k+1)^2(k+2)^2)}{4})\\ = (\cfrac{(k+1)(k+2))}{2})^2\\ = \left(1+2+...+k+(k+1)\right)^2$

proved by Induction.

$\therefore 1^3 + 2^3 + ....+ n^3 = (1+2+...+n)^2 = \left(\cfrac{10(11)}{2}\right)^2 = 55^2 = \boxed{3025}$

$1^3 + 2^3 + 3^3 + ... + n^3 = (1 + 2 + 3 + ... + n)^2 = \frac{n(n+1)}{2}^2$ , so $1^3 + 2^3 + 3^3 + ... + 10^3 = \frac{10 \times (10+1)}{2}^2 = 55^2 = \boxed{3025}$