已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q的等比数列.若a1=f(d-值1),a3=f(d+1),b1=f(q-1),b3=f(q+1) (Ⅰ)求数列an,bn的通项公式; (Ⅱ)设数列cn对任意自然数n均有c1/b1+c2/2b2+……+cn/nbn=an+1,
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![已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q的等比数列.若a1=f(d-值1),a3=f(d+1),b1=f(q-1),b3=f(q+1) (Ⅰ)求数列an,bn的通项公式; (Ⅱ)设数列cn对任意自然数n均有c1/b1+c2/2b2+……+cn/nbn=an+1,](/uploads/image/z/2084106-66-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%28x-1%29%5E2%2C%E6%95%B0%E5%88%97an%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BAd%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cbn%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BAq%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.%E8%8B%A5a1%3Df%28d-%E5%80%BC1%29%2Ca3%3Df%28d%2B1%29%2Cb1%3Df%28q-1%29%2Cb3%3Df%28q%2B1%29+%28%E2%85%A0%29%E6%B1%82%E6%95%B0%E5%88%97an%2Cbn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B+%28%E2%85%A1%29%E8%AE%BE%E6%95%B0%E5%88%97cn%E5%AF%B9%E4%BB%BB%E6%84%8F%E8%87%AA%E7%84%B6%E6%95%B0n%E5%9D%87%E6%9C%89c1%2Fb1%2Bc2%2F2b2%2B%E2%80%A6%E2%80%A6%2Bcn%2Fnbn%3Dan%2B1%2C)
已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q的等比数列.若a1=f(d-值1),a3=f(d+1),b1=f(q-1),b3=f(q+1) (Ⅰ)求数列an,bn的通项公式; (Ⅱ)设数列cn对任意自然数n均有c1/b1+c2/2b2+……+cn/nbn=an+1,
已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q的等比数列.若a1=f(d-值
1),a3=f(d+1),b1=f(q-1),b3=f(q+1) (Ⅰ)求数列an,bn的通项公式; (Ⅱ)设数列cn对任意自然数n均有c1/b1+c2/2b2+……+cn/nbn=an+1,求c1+c3+……+c(2n-1) 的和Tn
已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q的等比数列.若a1=f(d-值1),a3=f(d+1),b1=f(q-1),b3=f(q+1) (Ⅰ)求数列an,bn的通项公式; (Ⅱ)设数列cn对任意自然数n均有c1/b1+c2/2b2+……+cn/nbn=an+1,
a3-a1=2d=f(d+1)-f(d-1)
=d^2-(d-2)^2
=4d-4
所以2d=4----> d=2
a1=f(2-1)=f(1)=(1-1)^2=0
所以an=2*(n-1)
b3/b1=q^2=q^2/(q-2)^2
q-2=1或q-2=-1
得q1=3或q2=1
当q1=3时,b1=4---->得:bn=4*3^(n-1)
当q2=1时,b1=1---->得:bn=1是一个常数列
∵a1=f(d-1)=(d-2)2,a3=f(d+1)=d2,
∴a3-a1=d2-(d-2)2=2d,
∴d=2,∴an=a1+(n-1)d=2(n-1);又b1=f(q+1)=q2,b3=f(q-1)=(q-2)2,
∴=q2,由q∈R,且q≠1,得q=-2,
∴bn=b*qn-1=4*(-2)n-1