已知等差数列{an}的前n项和是Sn,且a2=1,S5=15.(1)求an(2)若数列{bn}满足a1b1+a2b2+a3b3+...+anbn=10+(2n-5)*2^(n+1),求bn
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![已知等差数列{an}的前n项和是Sn,且a2=1,S5=15.(1)求an(2)若数列{bn}满足a1b1+a2b2+a3b3+...+anbn=10+(2n-5)*2^(n+1),求bn](/uploads/image/z/9315459-27-9.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E6%98%AFSn%2C%E4%B8%94a2%3D1%2CS5%3D15.%EF%BC%881%EF%BC%89%E6%B1%82an%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3a1b1%2Ba2b2%2Ba3b3%2B...%2Banbn%3D10%2B%282n-5%29%2A2%5E%28n%2B1%29%2C%E6%B1%82bn)
已知等差数列{an}的前n项和是Sn,且a2=1,S5=15.(1)求an(2)若数列{bn}满足a1b1+a2b2+a3b3+...+anbn=10+(2n-5)*2^(n+1),求bn
已知等差数列{an}的前n项和是Sn,且a2=1,S5=15.
(1)求an
(2)若数列{bn}满足a1b1+a2b2+a3b3+...+anbn=10+(2n-5)*2^(n+1),求bn
已知等差数列{an}的前n项和是Sn,且a2=1,S5=15.(1)求an(2)若数列{bn}满足a1b1+a2b2+a3b3+...+anbn=10+(2n-5)*2^(n+1),求bn
有题设可以求出an=2n-3
a1b1+.+anbn=10+(2n-5)*2^(n+1)
a1b1+.+a(n-1)b(n-1)=10+(2n-7)*2^n
两式相减得出bn=2^n