- 设计一个算法求s=1的平方2的平方3**************?
- 一个算法求 S = 1²+2²+3²+...+100²
∵(n-1)³ = (n-1)(n²-2n+1) = n³-3n²+3n-1
--->n³-(n-1)³ = 3n²-3n+1，
令n分别 = 1、2、3、....、n--->
1³-0³ = 3•1² - 3•1 + 1
2³-1³ = 3•2² - 3•2 + 1
3³-2³ = 3•3² - 3•3 + 1
......
n³-(n-1)³=3•n²-3•n + 1
相加：n³ = 3(1²+2²+...+n²)-3(1+2+...+n)+n
= 3S - (3/2)n(n+1)+n
--->S = [n³-n+3n(n+1)/2]/3
= [2n(n-1)(n+1)+3n(n+1)]/6
= n(n+1)(2n+1)/6
令n=100--->S = 100×101×201÷6 = 338350