Question

数列题 急求

已知A1=1 A2=2 An，An+1，An+2在n为奇数时成等比，在n为偶数时成等差
求An通项

Answer 1

令k为正整数

a(2k-1),a(2k),a(2k+1)成等比数列
a(2k),a(2k+1),a(2k+2)成等差数列

a1=1
a2=2
a3=4
a4=6
a2^0.5=根号2
a4^0.5=2*根号2

a(2k-1)*a(2k+1)=a(2k)^2
a(2k)+a(2k+2)=2*a(2k+1)

可得

a(2k)=k*(k+1)
a(2k-1)=k*2

n为偶数时，an=(n/2)*(n/2+1)=n*(n+2)/4

n为奇数时，an=(n+1)^2/4

追问

a(2k-1)*a(2k+1)=a(2k)^2
a(2k)+a(2k+2)=2*a(2k+1)

怎么化简到下一步的

追答

a(2k-1)*a(2k+1)=a(2k)^2
a(2k)+a(2k+2)=2*a(2k+1)

(a(2k-1)*a(2k+1))^0.5+(a(2k+1)*a(2k+3))^0.5=2*a(2k+1)

(a(2k-1))^0.5+(a(2k+3))^0.5=2*(a(2k+1))^0.5

即(a(2k+1))^0.5构成等差数列

a1^0.5=1
a3^0,5=2
(a(2k-1))^0.5=k+1
a(2k-1)=k^2

a(2k)=(a(2k-1)*a(2k+1))^0.5=k*(k+1)