Question

求助一道高中数学题

1.已知数列｛An／n}的前N项和为Sn,且满足A1=1,An=A(n-1)+n,则Sn=?

Answer 1

An=A(n-1)+n =>an-a(n-1)=n
an=(an-a(n-1))+(a(n-1)-a(n-2))+...+(a3-a2)+(a2-a1)+a1
=n+(n-1)+...+3+2+1=n(n+1)/2
an/n=(n+1)/2是等差数列
sn=n(1+(n+1)/2)/2=n(n+3)/4

Answer 2

An=A(n-1)+n 移项可得：An-A(n-1)=n
再用累加法可得：
An=(An-A(n-1))+(A(n-1)-A(n-2))+...+(A3-A2)+(A2-A1)+a1
=n+(n-1)+...+3+2+1=n(n+1)/2
An/n=(n+1)/2是等差数列首项为1公差为二分之一
故而：sn=n(1+(n+1)/2)/2=n(n+3)/4

Answer 3

见回答1,2