Cn=anbn,求数列{an}的前n项和,bn=2n+1,an=2^n
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Cn=anbn,求数列{an}的前n项和,bn=2n+1,an=2^n
![Cn=anbn,求数列{an}的前n项和,bn=2n+1,an=2^n](/uploads/image/z/19466898-42-8.jpg?t=Cn%3Danbn%2C%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Cbn%3D2n%2B1%2Can%3D2%5En)
cn=anbn=(2n+1)×2ⁿ=n×2^(n+1)+2ⁿ
Sn=c1+c2+...+cn=[1×2²+2×2³+...+n×2^(n+1)] +(2+2²+...+2ⁿ)
令Cn=1×2²+2×2³+...+n×2^(n+1)
则2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1)-n×2^(n+2)
Cn=n×2^(n+2)-[2²+2³+...+2^(n+1)]
Sn=Cn+(2+2²+...+2ⁿ)
=n×2^(n+2)-[2²+2³+...+2^(n+1)]+(2+2²+...+2ⁿ)
=n×2^(n+2)+2 -2^(n+1)
=(2n-1)×2^(n+1) +2
Sn=c1+c2+...+cn=[1×2²+2×2³+...+n×2^(n+1)] +(2+2²+...+2ⁿ)
令Cn=1×2²+2×2³+...+n×2^(n+1)
则2Cn=1×2³+2×2⁴+...+(n-1)×2^(n+1)+n×2^(n+2)
Cn-2Cn=-Cn=2²+2³+...+2^(n+1)-n×2^(n+2)
Cn=n×2^(n+2)-[2²+2³+...+2^(n+1)]
Sn=Cn+(2+2²+...+2ⁿ)
=n×2^(n+2)-[2²+2³+...+2^(n+1)]+(2+2²+...+2ⁿ)
=n×2^(n+2)+2 -2^(n+1)
=(2n-1)×2^(n+1) +2
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