函数f(x)=(x-1)²,数列(an)为等差数列,公差为d,数列(bn)为等比数列,公比为q(q≠0,q≠1).设a1=f(d-1),a2=f(d+1),b1=f(g-1),b2=f(g+1).
问题描述:
函数f(x)=(x-1)²,数列(an)为等差数列,公差为d,数列(bn)为等比数列,公比为q(q≠0,q≠1).设a1=f(d-1),a2=f(d+1),b1=f(g-1),b2=f(g+1).
(1)求数列{an} {bn}的通项公式;
(2)设对任意自然数n,都有a(n+1)=c1/b1+c2/b2+…+cn/bn,求c1+c2+c3+…cn.
(注:数列a,b后面都是角标)
答
(1) d=a2-a1=f(d+1)-f(d-1)=d^2-(d-2)^2=2(2d-2)d=4d-4d=4/3a1=f(d-1)=f(1/3)=(1/3-1)^2=4/9所以 an=a1+(n-1)d=4/9+(n-1)*4/3=4n/3-8/9q=b2/b2=f(q+1)/f(q-1)=q^2/(q-2)^2q=(q-2)^2q^2-5q+4=0q=4 或1(舍去)b1=f(q-1)...