已知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,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)求{an}{bn}的通项公式设数列{cn}满足一切n∈N*,都有c1/b1+c2/b2+……+cn/bn=a(n+1下标)成立,求{cn}
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![已知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,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)求{an}{bn}的通项公式设数列{cn}满足一切n∈N*,都有c1/b1+c2/b2+……+cn/bn=a(n+1下标)成立,求{cn}](/uploads/image/z/3622937-41-7.jpg?t=%E5%B7%B2%E7%9F%A5f%28x%29%3D%28x-1%29%5E2%2C%E6%95%B0%E5%88%97%7Ban%7D%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%2C%7Bbn%7D%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%2C%E8%AE%BEa1%3Df%28d-1%29%2Ca3%3Df%28d%2B1%29%2Cb1%3Df%28q%2B1%29%2Cb3%3Df%28q-1%29%E6%B1%82%7Ban%7D%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E8%AE%BE%E6%95%B0%E5%88%97%7Bcn%7D%E6%BB%A1%E8%B6%B3%E4%B8%80%E5%88%87n%E2%88%88N%2A%2C%E9%83%BD%E6%9C%89c1%2Fb1%2Bc2%2Fb2%2B%E2%80%A6%E2%80%A6%2Bcn%2Fbn%3Da%28n%2B1%E4%B8%8B%E6%A0%87%EF%BC%89%E6%88%90%E7%AB%8B%2C%E6%B1%82%7Bcn%7D)
已知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,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)求{an}{bn}的通项公式设数列{cn}满足一切n∈N*,都有c1/b1+c2/b2+……+cn/bn=a(n+1下标)成立,求{cn}
已知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,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)
求{an}{bn}的通项公式
设数列{cn}满足一切n∈N*,都有c1/b1+c2/b2+……+cn/bn=a(n+1下标)成立,求{cn}通项公式
*第一题可以不答,我会了~
已知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,设a1=f(d-1),a3=f(d+1),b1=f(q+1),b3=f(q-1)求{an}{bn}的通项公式设数列{cn}满足一切n∈N*,都有c1/b1+c2/b2+……+cn/bn=a(n+1下标)成立,求{cn}
那太简单啦,通过递推不就得了,你第一问求得的是d=2,q=3吧.
由原式再往上递推一项,就有c1/b1+c2/b2+……+cn/bn+c(n+1)/b(n+1)=a(n+2),
然后跟原式联立,两式相减,不就得c(n+1)/b(n+1)=2了嘛,结果就是cn=2bn.
以后看到那么复杂的加和等于什么的,都可以尝试通过递推式解决,化繁为简.