数列{an}的前n项和为Sn且Sn=n(n+1) 1 若数列{bn}满足an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)求{bn}通项公式2 令cn=anbn/4求数列{cn}的前n项和Tn
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![数列{an}的前n项和为Sn且Sn=n(n+1) 1 若数列{bn}满足an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)求{bn}通项公式2 令cn=anbn/4求数列{cn}的前n项和Tn](/uploads/image/z/6130911-39-1.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%E4%B8%94Sn%3Dn%28n%2B1%29+1+%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3an%3Db1%2F%283%2B1%29%2Bb2%2F%283%5E2%2B1%29%2Bb3%2F%283%5E3%2B1%29%2B%E2%80%A6%E2%80%A6bn%2F%283%5En%2B1%29%E6%B1%82%7Bbn%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F2+%E4%BB%A4cn%3Danbn%2F4%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bcn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
数列{an}的前n项和为Sn且Sn=n(n+1) 1 若数列{bn}满足an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)求{bn}通项公式2 令cn=anbn/4求数列{cn}的前n项和Tn
数列{an}的前n项和为Sn且Sn=n(n+1) 1 若数列{bn}满足an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)
求{bn}通项公式
2 令cn=anbn/4求数列{cn}的前n项和Tn
数列{an}的前n项和为Sn且Sn=n(n+1) 1 若数列{bn}满足an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)求{bn}通项公式2 令cn=anbn/4求数列{cn}的前n项和Tn
(1)an=Sn-S(n-1)=n(n+1)-(n-1)n=2n
因为an=b1/(3+1)+b2/(3^2+1)+b3/(3^3+1)+……bn/(3^n+1)
所以an-a(n-1)=bn/(3^n+1)
而an-a(n-1)=2
所以bn/(3^n+1)=2 得到bn=2(3^n+1)
(2)cn=anbn/4=n(3^n+1)=n×3^n+n;
令dn=n×3^n Pn是dn的前n项和;
那么Pn=1×3+2×3^2+...+n×3^n ①
所以3Pn=1×3^2+2×3^3+...+n×3^(n+1) ②
②-①得到 2Pn=n×3^(n+1)-(3+3^2+...+3^n)=n×3^(n+1)-[3^(n+1)-3)/2]
化简后Pn=[(2n-1)×3^(n+1)+3]/4;
所以Tn=Pn+n(n+1)/2=[(2n-1)×3^(n+1)+2n^2+2n+3]/4;
数列{an}的前n项和为Sn且Sn=n(n+1),
易知an=Sn-S
待续