作业帮高一数学题,求详细解答
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高一数学题,求详细解答
1.数列{an}是等差数列,a1=f（x+1）,a2=0,a3=f（x-1）,其中f（x）=x^2-4x+2,求通项公式.
2.在以d为公差的等差数列{an}中,设S1=a1+a2··an,S2=（an+）+（an+2）··a2n,S3=（a2n+1)+(a2n+2)··a3n.求证S1,S2,S3也是等差数列,并求公差
1.数列{an}是等差数列,a1=f（x+1）,a2=0,a3=f（x-1）,其中f（x）=x^2-4x+2,求通项公式.
2.在以d为公差的等差数列{an}中,设S1=a1+a2··an,S2=（an+）+（an+2）··a2n,S3=（a2n+1)+(a2n+2)··a3n.求证S1,S2,S3也是等差数列,并求公差
f(x) = (x-2)^2 - 2 ,a1 + a3 = 2a2 = 0 ,(x-1)^2 - 2 + (x-3)^2 - 2=0,
解得：x = 1或3
x = 1时,a1 = -2 ,d = 2 ,an = -2 + 2(n-1) = 2n-4 ,
x = 3时,a1 = 2 ,d = -2 ,an = 2 - 2(n-1) = -2n+4 ,
S2 - S1 = [a(n+1) - a1] + [a(n+2) - a2] + ...+ [a(2n) - an] = nd·n =
d·n^2 ,同理可得S3 - S2 = d·n^2 = S2 - S1 ,所以S1,S2,S3也是等差数列 ,公差 = d·n^2
解得：x = 1或3
x = 1时,a1 = -2 ,d = 2 ,an = -2 + 2(n-1) = 2n-4 ,
x = 3时,a1 = 2 ,d = -2 ,an = 2 - 2(n-1) = -2n+4 ,
S2 - S1 = [a(n+1) - a1] + [a(n+2) - a2] + ...+ [a(2n) - an] = nd·n =
d·n^2 ,同理可得S3 - S2 = d·n^2 = S2 - S1 ,所以S1,S2,S3也是等差数列 ,公差 = d·n^2