(高一数学)已知f(x+1)=x的平方减2，等差数列(an)中，a1=f(x－1)，a2=－3/2,a3=f(x).

(1)f(x+1)=x^2-2
令x+1=t x=t-1
f(t)=(t-1)^2-2
即f(x)=(x-1)^2-2
a1=f(x-1)=(x-2)^2-2
a3=f(x)=(x-1)^2-2
等差数列{an}
所以a1+a3=-2a2=-3
2x^2-6x+4=0
x^3-3x+2=0
x=1 or x=2
(2)
x=1时 a1=f(0)=-1 a3=f(1)=-2
公差为-1/2
通项公式an=-1-1/2(n-1)=-(n+1)/2=-1/2*(n+1)
a2+a5+a8+``````+a26
=-1/2(3+6+9+.....27)
=[-1/2][(3+27)*9/2]
=-135/2

x=2时 a1=f(1)=-2 a3=f(2)=-1
公差为1/2
通项公式an=-2+1/2(n-1)=(n-5)/2=1/2*(n-5)
a2+a5+a8+``````+a26
=1/2(-3+0+3+....+21)
=[1/2][(-3+21)*9/2]
=81/2

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