已知:an=3n-1,bn=2^n,求数列{anbn}的前n项和
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已知:an=3n-1,bn=2^n,求数列{anbn}的前n项和
cn=anbn=(3n-1)*2^n
Sn=2*2^1+5*2^2+……+(3n-1)*2^n
2Sn= 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+1)+12-4
=(3n-4)*2^(n+1)+8
Sn=2*2^1+5*2^2+……+(3n-1)*2^n
2Sn= 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+1)+12-4
=(3n-4)*2^(n+1)+8
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