问题
解答题
已知函数f(x)=e-x(cosx+sinx),将满足f'(x)=0的所有正数x从小到大排成数列{xn}. (Ⅰ)证明数列{f{xn}}为等比数列; (Ⅱ)记Sn是数列{xnf{xn}}的前n项和,求
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答案
(Ⅰ)证明:f'(x)=-e-x(cosx+sinx)+e-x(-sinx+cosx)=-2e-xsinx.
由f'(x)=0,得-2e-xsinx=0.
解出x=nπ,n为整数,从而xn=nπ,n=1,2,3,f(xn)=(-1)ne-nπ.
=-e-π.f(xn+1) f(xn)
所以数列{f{xn}}是公比q=-e-π的等比数列,且首项f(x1)=q.
(Ⅱ)Sn=x1f(x1)+x2f(x2)++xnf(xn)=πq(1+2q++nqn-1),
qSn=πq(q+2q2++nqn),
Sn-qSn=πq(1+2q2++qn-1-nqn)
=πq(
-nqn),1-qn 1-q
从而S1+S2++Sn n
=
-πq (1-q)2
(1+q++qn-1)-πq2 n(1-q)2
(1+2q++nqn-1)πq2 n(1-q)
=
-πq (1-q)2 πq2 n(1-q)2
-1-qn 1-q
(πq2 n(1-q)2
-nqn)1-qn 1-q
=
-πq (1-q)2
(1-qn)+2πq2 n(1-q)3
.πqn+2 (1-q)2
因为|q|=e-π<1.
qn=0,lim n→∞
所以lim n→∞
=S1+S2++Sn n
=πq (1-q)2
.-πeπ (eπ+1)2