Question

已知等差数列{an}的前n项和为Sn,若Sn=S（n-1）+n+2(n属于正整数,n大于等于2),则S5=多少

求助啊···········

Answer 1

s(n+1)=s(n)+(n+1)+2=s(n)+[(n+1)(n+2)-n(n+1)]/2 + 2[(n+1)-n],
s(n+1) - (n+1)(n+2)/2 - 2(n+1) = s(n) - n(n+1)/2 - 2n,
{s(n)-n(n+1)/2 - 2n}是首项为s(1)-1-2=s(1)-3的常数列。
s(n)-n(n+1)/2-2n = s(1)-3,
s(n) = n(n+1)/2 + 2n + s(1)-3,
s(5) = 5*3 + 10 + s(1) - 3 = 22 + s(1) = 22 + a(1)

Answer 2

Sn=S（n-1）=an=n+2,等差数列，再用公式就行