已知{an}满足an=(3n+1)2^(n+1),求Sn,用错位相减法
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![已知{an}满足an=(3n+1)2^(n+1),求Sn,用错位相减法](/uploads/image/z/2708636-68-6.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E6%BB%A1%E8%B6%B3an%3D%283n%2B1%292%5E%EF%BC%88n%2B1%EF%BC%89%2C%E6%B1%82Sn%2C%E7%94%A8%E9%94%99%E4%BD%8D%E7%9B%B8%E5%87%8F%E6%B3%95)
已知{an}满足an=(3n+1)2^(n+1),求Sn,用错位相减法
已知{an}满足an=(3n+1)2^(n+1),求Sn,用错位相减法
已知{an}满足an=(3n+1)2^(n+1),求Sn,用错位相减法
依题意得
Sn=4×2^2+7×2^3+10×2^4+…+(3n+1)×2^(n+1)
所以等式左右乘以等比数列的公比2,得:
2Sn=4×2^3+7×2^4+10×2^5+…+(3n+1)×2^(n+2)
所以2Sn-Sn=Sn=-4×2^2-[3×2^3+3×2^4+…+3×2^(n+1)]+(3n+1)×2^(n+2)
=-16-3[2^3+2^4+2^5+…+2^(n+1)]+(3n+1)×2^(n+2)
=-16-3[-2^3+2^(n+2)]+(3n+1)×2^(n+2)
整理得
=(3n-2)×2^(n+2)+8
已验算
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