- 数学问题6在以d为公差的等差数列{an}中,设S1=a1+a2+
- 在以d为公差的等差数列{an}中,设S1=a1+a2+…an ,S2=an+1(n+1在a的右下角)
S2=an+1(n+1在a的右下角) +an+2(n+2在a的右下角)+…+a2n(2n在a的右下角)
S3=a2n+1(2n+1在a的右下角)+ a2n+2(2n+2在a的右下角)+…a3n(3n在a的右下角)
求证S1,S2,S3也是等差数列,并求其公差
数列{an}是等差数列,a1=f(x+1),a2=0 a3=f(x-1)其中f(x)=x的平方-4x+2 求通项公式an
- (1)
S3-S2=n*nd=n^2d
S2-S1=n*nd=n^2d
所以S1,S2,S3成等差数列,公差是n^2d
(2)
a1+a3=2a2=0
f(x+1)+f(x-1)
=(x+1)^2-4(x+1)+2+(x-1)^2-4(x-1)+2=2x^2-8x+6=0
x=1或x=3
当x=1时,a1=f(2)=-2, d=a2-a1=2, 通项公式an=-2+2(n-1)=2n-4
当x=3时,a1=f(4)=2, d=a2-a1=-2, 通项公式an=2-2(n-1)=4-2n