有穷数列{an}共有2k项(整数k≥2),a1=2,a(n+1)=(a-1)Sn+2(n=1,2,...,2k-1),其中常数a>1 求:若a=2^(2/(2k-1)),数列{bn}满足bn=(1/n)x log2(a1a2...an),(n=1,2,...,2k),求数列{bn}的通项公式
问题描述:
有穷数列{an}共有2k项(整数k≥2),a1=2,a(n+1)=(a-1)Sn+2(n=1,2,...,2k-1),其中常数a>1 求:
若a=2^(2/(2k-1)),数列{bn}满足bn=(1/n)x log2(a1a2...an),(n=1,2,...,2k),求数列{bn}的通项公式
答
bn=1/n[n+n(n-1) /(2k-1 )] ]=(n-1)/(2k-1)+1(n=1,2,……2k).
答
当n=1时,a2=2a,a2/a1=a;当2≤n≤2k-1时,an+1=(a-1)Sn+2,an=(a-1)Sn-1+2∴an+1-an=(a-1)an∴an+1/an=a∴数列{an}是首项为2,公比为a的等比数列∴an=2a^n-1 又a=2^[2/(2k-1)]∴a1×a2×…an=2^na^[1+2+…+(n-1...