an=(n+2)2^n,求Sn,用错位相减法

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an=(n+2)2^n,求Sn,用错位相减法an=(n+2)2^n,求Sn,用错位相减法an=(n+2)2^n,求Sn,用错位相减法依题意得Sn=4×2^2+7×2^3+10×2^4+…+(3n+1)

an=(n+2)2^n,求Sn,用错位相减法
an=(n+2)2^n,求Sn,用错位相减法

an=(n+2)2^n,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