s'il vous plait tu peux maider bayin bi l istidlal bi tarajo3 al 3ibara 1³+2³+·····+n³=n²(n+1)² /4

[tex]S_n = \frac{n^2(n+1)^2}{4} \\ \\ S_1 = \frac{1^2 (1+1)^2}{4} = \frac{4}{4} = 1 = 1^3 \\ \\ S_2 = \frac{2^2(2+1)^2}{4} = \frac{4*9}{4} = 9 = 1^3+2^3 \\ \\ S_{n-1} = \frac{(n-1)^2(n-1+1)^2}{4} = \frac{(n-1)^2n^2}{4} \\ \\ S_{n-1} + n^3 = \frac{(n-1)^2n^2}{4}+n^3 = \frac{(n-1)^2n^2+4nn^2}{4} \\ \\ = \frac{n^2[(n-1)^2+4n]}{4} = \frac{n^2(n^2-2n+1+4n)}{4} \\ \\ = \frac{n^2(n+1)^2}{4} = S_{n} \\ [/tex]

Il est prouvé par induction