Question

数列{an}中 a1=1 a(n+1)=an+2n (n≥2)求an

如何用累加法做 求过程

Answer 1

累加法：
a(n) = a(n-1) + 2(n-1),
a(n-1) = a(n-2) + 2(n-2),
...
a(3) = a(2) + 2*2
a(2) = a(1) + 2*1,

a(n) + a(n-1) + ... + a(3) + a(2) = a(n-1) + a(n-2) + ... + a(2) + a(1) + 2[1+2+...+(n-1)],
a(n) = a(1) + 2(n-1)n/2 = 1 + (n-1)n.

我推荐如下解法：
a(n+1) = a(n) + 2n = a(n) + n(n+1) - (n-1)n,
a(n+1) - n(n+1) = a(n) - (n-1)n,
{a(n) - (n-1)n}是首项为a(1)=1,的常数数列。
a(n) - (n-1)n = 1,
a(n) = 1 + (n-1)n.