已知：an=3n-1,bn=2^n,求数列{anbn}的前n项和
人气：286 ℃ 时间：2019-08-26 06:40:48
解答
cn=anbn=(3n-1)*2^nSn=2*2^1+5*2^2+……+(3n-1)*2^n2Sn= 2*2^2+……+(3n-4)*2^n+(3n-1)*2^(n+1)相减：Sn=(3n-1)*2^(n+1)-3*(2^2+2^3+……+2^n)-2*2^1=(3n-1)*2^(n+1)-3*[2^2-2^(n+1)]/(1-2)-4=(3n-1)*2^(n+1)-3*2^(n...
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