设数列{an}满足当n=2k-1(k∈N﹢)时 ,an=n,当;当n=2k(k∈N*)时,an=ak.,记Sn=a1+a2+a3……+a2n-1+a2n(1)求S3 (2)证明Sn=4^n-1+Sn-1(n>=2)证明1/S1+1/S2+……+Sn-1
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设数列{an}满足当n=2k-1(k∈N﹢)时 ,an=n,当;当n=2k(k∈N*)时,an=ak.,记Sn=a1+a2+a3……+a2n-1+a2n(1)求S3 (2)证明Sn=4^n-1+Sn-1(n>=2)证明1/S1+1/S2+……+Sn-1
设数列{an}满足当n=2k-1(k∈N﹢)时 ,an=n,当;当n=2k(k∈N*)时,an=ak.,记Sn=a1+a2+a3……+a2n-1+a2n(1)求S3 (2)证明Sn=4^n-1+Sn-1(n>=2)
证明1/S1+1/S2+……+Sn-1
设数列{an}满足当n=2k-1(k∈N﹢)时 ,an=n,当;当n=2k(k∈N*)时,an=ak.,记Sn=a1+a2+a3……+a2n-1+a2n(1)求S3 (2)证明Sn=4^n-1+Sn-1(n>=2)证明1/S1+1/S2+……+Sn-1
证明如下:
(1)S3=a1+a2+a3+a4+a5+a6+a7+a8=a1+a1+a3+a1+a5+a3+a7+a1
=4a1+2a3+a5+a7=4×1+2×3+5+7=22
(2)Sn=a1+a2+…+a2n-1+a2n
=(a1+a3+…+a2n-1)+(a2+a4+…+a2n)
=[1+3+…+(2n-1)]+(a2+a4+a6+…+a2n)
=4n-1+(a1+a2+a3+…+a2n-1)
=4^(n-1)+Sn-1
(3)由(2)知Sn-Sn-1=4^(n-1),于是有:Sn-1-Sn-2=4^(n-2),Sn-2-Sn-3=4^(n-3) …s2-s1=4
上述各式相加得:Sn-S1=4+4^2+…+4^(n-1)
sn=2+4(1-4^(n-1))/(1-4)=1/3*(2+4^n),
∴1/sn=3/(4^n+2)<3/4^n
∴1/s1+1/s2+1/s3+…+1/sn<3/4(1+1/4+1/4^2+…+1/4^(n-1))=1-1/4^n
原式得证