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](/uploads/image/z/8829050-50-0.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%E5%BD%93n%3D2k-1%EF%BC%88k%E2%88%88N%EF%B9%A2%EF%BC%89%E6%97%B6+%2Can%3Dn%2C%E5%BD%93%EF%BC%9B%E5%BD%93n%3D2k%EF%BC%88k%E2%88%88N%2A%EF%BC%89%E6%97%B6%2Can%3Dak%EF%BC%8E%2C%E8%AE%B0Sn%3Da1%2Ba2%2Ba3%E2%80%A6%E2%80%A6%2Ba2n-1%2Ba2n%EF%BC%881%EF%BC%89%E6%B1%82S3+%EF%BC%882%EF%BC%89%E8%AF%81%E6%98%8ESn%3D4%5En-1%2BSn-1%28n%3E%3D2%29%E8%AF%81%E6%98%8E1%2FS1%2B1%2FS2%2B%E2%80%A6%E2%80%A6%2BSn-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
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