an=n*2^n,求Sn

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an=n*2^n,求Snan=n*2^n,求Snan=n*2^n,求SnSn=1×2＋2×2²＋3×2³＋…＋n×2^n2Sn=1×2²＋2×2³＋3×2^4＋

an=n*2^n,求Sn
an=n*2^n,求Sn

an=n*2^n,求Sn
Sn=1×2＋2×2²＋3×2³＋…＋n×2^n
2Sn=1×2²＋2×2³＋3×2^4＋…＋(n－1)×2^n＋n×2^(n＋1)
两式相减,得：
－Sn=2＋2²＋2³＋…＋2^n－n×2^(n＋1)
=[2^(n＋1)－2]－n×2^(n＋1)
=(1－n)×2^(n＋1)－2
所以,Sn=(n－1)×2^(n＋1)＋2.

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