Question

等比数列【An】中,前n项和为Sn,已知S1,S2,S3成等差数列

Answer 1

S1=a1

S2=a1(1+q)

S3=a1(1+q+q^2)

S1,S3,S2成等差数列

即

s3-s1=s2-s3

1+q+q^2-1=1+q-(1+q+q^2)

q^2+q=-q^2

q=0,或-1/2

如果a1-a3=3

a1不等于a3

q不等于0,即q=-1/2

a1(1-1/4)=3

a1=4

所以 an=4*(-1/2)^(n-1)

bn=2n+4*(-1/2)^(n-1)=2n+4*(-1/2)^n*(-2)=2n-8*(-1/2)^n

Tn=2(1+2+.+n)-8[(-1/2)+(-1/2)^2+...+(-1/2)^n]

=n(1+n)-8*(-1/2)*(1-( -1/2)^n)/(1+1/2)

=n(1+n)+8/3*[1-(-1/2)^n]