1^3+2^3+3^3+....+n^3=1/4n^2(n+1)^2

1³ + 2³ + 3³ + ... + n³ = (n²(n+1)²)/4

Untuk
n = 1
1³ = (1²(1+1)²)/4
1 = (1(4))/4
1 = 1
*untuk n = 1 bernilai benar*

anggap benar untuk n = k
1³ + 2³ + 3³ + ... + k³ = (k²(k+1)²)/4

untuk n = k + 1

1³ + 2³ + 3³ + .. + k³ + (k+1)³
= [tex] \frac{(k^2(k+1)^2}{4} + (k+1)^3 [/tex]
= [tex] \frac{k^2(k+1)^2}{4} [/tex] + [tex] \frac{4(k+1)(k+1)^2}{4} [/tex]
= [tex] \frac{(k+1)^2(k^2+4(k+1))}{4} [/tex]
= [tex] \frac{(k+1)^2(k^2+4k+1)}{4} [/tex]
= [tex] \frac{(k+1)^2(k+2)^2}{4} [/tex]
= [tex] \frac{(n)^2(n+1)^2}{4} [/tex]
*terbukti*