Solusi

Kita akan buktikan dengan induksi matematika

Untuk $k=1$
$1^3=1=\large{(\frac{1.2}{2})^2}$

Asumsikan benar untuk $k=n$
$1^3+2^3+3^3+...+n^3=\large{(\frac{n(n+1)}{2})^2}$

Buktikan benar untuk $k=n+1$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{(\frac{n(n+1)}{2})^2}+(n+1)^3$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{n^2(n^2+2n+1)}{4}+\frac{4(n^3+3n^2+3n+1)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{n^2(n^2+2n+1)+4(n^3+3n^2+3n+1)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{(n^4+2n^3+n^2)+(4n^3+12n^2+12n+4)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{n^4+6n^3+13n^2+12n+4}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{(n^4+4n^3+4n^2)+(2n^3+8n^2+8n)+(n^2+4n+4)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{n^2(n^2+4n+4)+2n(n^2+4n+4)+(n^2+4n+4)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{(n^2+2n+1)(n^2+4n+4)}{4}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{\frac{(n+1)^2(n+2)^2}{2^2}}$

$1^3+2^3+3^3+...+n^3+(n+1)^3=\large{(\frac{(n+1)(n+2)}{2})^2}$

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