Question

1>等差数列｛an｝中,a＞0,S4=S9,则S取最大值时,n=

Answer 1

1. 是a1>0吧 S4=S9

=>a1+a2+a3+a4=a1+a2+...+a9

=>a7=0 所以S(max)=S6=S7

2. Sn=a1+a2+...+an

=(2^1+2*1-1)+(2^2+2*2-1)+(2^3+2*3-1)+...+(2^n+2n-1)

=2^1+2^2+...+2^n+2(1+2+3+...+n)-(1+1+...+1)

=2(1-2^n)/(1-2)+n(n+1)-n =2^(n+1)-2+n^2

3.an=1/[√n+√(n+1)]=√(n+1)-√n

Sn=a1+a2+...+an =√2-√1+√3-√2+...+√(n+1)-√n

=√(n+1)-1 Sn=√(n+1)-1=10 => n=120