[TEX]S_{n+1} - S_{n} = (n+1)^3 = n^3 + 3.n^2 + 3.n +1 \\ S_n = C \\ S_n = An^4 + B.n^3 + C.n^2 + D.n \\ S_{n+1} - S_{n} = A.(n+1)^4 +B.(n+1)^3 + C.(n+1)^2 + D.(n+1) - An^4 - B.n^3 - C.n^2 - D.n = n^3 + 3.n^2 + 3.n +1\\ 4.A.n^3 + (6A +3B).n^2 +(4A +3B +2C)n + (A+B+C+D) = n^3 + 3.n^2 + 3.n +1 \\ A = \frac{1}{4}, B = \frac{1}{2}, C = \frac{1}{4}, D = 0 \\ S_n = C + 1/4.n^4 +1/2.n^3 + 1/4.n^2 \\ S_1 = 1 \Rightarrow C = 0 \\ S_n = \frac{1}{4}.n^2 .(n+1)^2[/TEX]
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[TEX] S_n = \frac{n^2 .(n+1)^2}{4}[/TEX]