问题 解答题

已知数列{an}是首项为a1=4,公比q≠1的等比数列,Sn是其前项和,且4a1,a5,-2a3成等差数列.

(1)求公比q的值;

(2)设An=S1+S2+S3+…+Sn,求An

答案

(1)∵4a1,a5,-2a3成等差数列,

∴2a5=4a1+(-2a3),

∵a5=a1q4,a3=a1q2

∴2a1q4=4a1-2•a1q2

∵a1≠0

q4+q2-2=0.

∴q2=1或q2=-2( 舍去)

∵q≠1,

∴q=-1.

(2)∵Sn=

4[1-(-1)n]
1-(-1)
=2-2•(-1)n

∴An=S1+S2+S3+…+Sn=[2-2•(-1)1]+[2-2•(-1)2]+[2-2•(-1)3]+…+[2-2•(-1)n]

=2n-2•[(-1)+(-1)2+(-1)3+…+(-1)n]

=2n-2

-1•[1-(-1)n]
1-(-1)
=2n+1-(-1)n

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