已知数列{an}满足an+1=qan+2q-2,(q为常数),若a3,a4,a5,a6∈{﹣18,﹣6,6,30},则a1=
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![已知数列{an}满足an+1=qan+2q-2,(q为常数),若a3,a4,a5,a6∈{﹣18,﹣6,6,30},则a1=](/uploads/image/z/10340408-56-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3an%2B1%3Dqan%2B2q-2%2C%28q%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%89%2C%E8%8B%A5a3%2Ca4%2Ca5%2Ca6%E2%88%88%7B%EF%B9%A318%2C%EF%B9%A36%2C6%2C30%7D%2C%E5%88%99a1%3D)
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已知数列{an}满足an+1=qan+2q-2,(q为常数),若a3,a4,a5,a6∈{﹣18,﹣6,6,30},则a1=
已知数列{an}满足an+1=qan+2q-2,(q为常数),若a3,a4,a5,a6∈{﹣18,﹣6,6,30},则a1=
已知数列{an}满足an+1=qan+2q-2,(q为常数),若a3,a4,a5,a6∈{﹣18,﹣6,6,30},则a1=
a6=q*a5+2q-2,
a5=q*a4+2q-2,
a4=q*a3+2q-2,
a3=q*a2+2q-2,
所以a6+2=q(a5+2),
a5+2=q(a4+2),
a4+2=q(a3+2),
即a3+2,a4+2,a5+2,a6+2成等比数列,
因a3,a4,a5,a6∈(-18,-6,6,30),
所以a3+2,a4+2,a5+2,a6+2∈(-16,-4,8,32),
q=-2,a3+2=-4,a4+2=8,a5+2=-16,a6+2=32或q=-1/2,a3+2=32,a4+2=-16,a5+2=8,a6+2=-4
即q=-2,a3=-6,a4=6,a5=-18,a6=30或q=-1/2,a3=30,a4=-18,a5=6,a6=-6.
a2+2=(a3+2)/q=2,a2=0,a1+2=(a2+2)/q=-1,a1=-3
或a2+2=(a3+2)/q=-64,a2=-66,a1+2=(a2+2)/q=128,a1=126
所以a1=-3或126.
你这道题有问题啊,移向后能够直接解出
an=(3+2q)/(q-1),q是常数an也应该是常数,你看看是不是输入的时候输错了
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