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=
=2-2•(-1)n.4[1-(-1)n] 1-(-1)
∴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
=2n+1-(-1)n-1•[1-(-1)n] 1-(-1)