Средства и методы разработки

Такой еще метод определим S2(n) = sum(k^2), 0 <= k <= n S3(n) = sum(k^3), 0 <= k <= n тогда S3(n+1) = S3(n) + (n+1)^3 = sum((k+1)^3), 0 <= k <= n S3(n+1) = sum(k^3 + 3k^2 + 3k + 1) = S3(n) + 3*S2(n) + 3/2*n*(n+1) + (n+1) 3*S2(n) = (n+1)^3 - 3/2*n*(n+1) - (n+1) = (n+1)(n^2 + 2n + 1 - 3/2*n - 1) = (n+1)(n + 1/2)n S2(n) = n/3(n+1)(n+1/2)