Question

高中等差数列

{an}为等差数列，若a3+a5+a7+a9+a11=100 . 求3a9-a13 , S13

Answer 1

由A1=S1可得A1=1/2(A1+1/A1) 又因为A10 求出A1=1 当然S1=1
因为An=Sn-Sn-1
所以Sn=1/2(An+1/An)
Sn=1/2×[Sn-Sn-1+1/(Sn-Sn-1)]
==Sn=1/2×[(Sn-Sn-1)2+1]/(Sn-Sn-1)
==2Sn(Sn-Sn-1)=(Sn-Sn-1)2+1
==2S2n-2Sn*Sn-1=S2n-2Sn*Sn-1+S2n-1+1
==2S2n=S2n+S2n-1+1
==S2n-S2n-1=1
所列数列{S2n}是一个以S21=1为首项 1为公差的等差数列
所以S2n=1+(n-1)*1=n ==Sn=√n
所以An=Sn-Sn-1=√n=√(n-1)
An=√n-√(n-1)也满足A1=1
所以An=√n-√(n-1)

Answer 2

a(n+m)=an+am
a3+a5+a7+a9+a11=100得5a7=100
a7=a1+6d=20
3a9-a13=3(a1+8d)-(a1+12d)=2(a1+6d)=40
S13=(a1+a13)+(a2+a12)+....+a7=13a7=260

Answer 3

a3+a5+a7+a9+a11=100
5a7=100
a7=20
3a9-a13=3(a7+2d)-(a7+6d)=2a7=40
s13=a1+a2+....+a13=13a7=260