Giải thích các bước giải:
Chứng minh $1^3+2^3+3^3+...+n^3=\left(1+2+3+..+n\right)^2$
Với $n=1\to 1^3=1^2$ đúng
Với $n=2\to 1^3+2^3=\left(1+2\right)^2$ đúng
...
Giả sử $n=k$ đúng
$\to 1^3+2^3+...+k^3=\left(1+2+...+k\right)^2$ đúng
Ta cần chứng minh $n=k+1$ đúng
Ta có $1^3+2^3+..+k^3+\left(k+1\right)^3=\left(1+2+..+k\right)^3+\left(k+1\right)^3=\left(\dfrac{k\left(k+1\right)}{2}\right)^2+\left(k+1\right)^3$
$\to 1^3+2^3+..+k^3+\left(k+1\right)^3=\left(k+1\right)^2\left(\dfrac{k^2}{4}+k+1\right)$
$\to 1^3+2^3+..+k^3+\left(k+1\right)^3=\left(k+1\right)^2\left(\dfrac{k^2+4k+4}{4}\right)$
$\to 1^3+2^3+..+k^3+\left(k+1\right)^3=\left(k+1\right)^2.\dfrac{\left(k+2\right)^2}{4}$
$\to 1^3+2^3+..+k^3+\left(k+1\right)^3=\left(\dfrac{\left(k+1\right)\left(k+2\right)}{2}\right)^2$
$\to 1^3+2^3+..+k^3+\left(k+1\right)^3=\left(1+2+3+..+k+\left(k+1\right)\right)^2$
$\to n=k+1$ đúng
$\to 1^3+2^3+...+n^3=\left(1+2+3+..+n\right)^2$
$\to \left(1^3+2^3+3^3+..+100^3\right)=\left(1+2+3+..+100\right)^2\quad\vdots\quad \left(1+2+3+..+100\right)$
$\to đpcm$
