(1)Sn=2an+1Sn-1=2an-1+1所以an=2an-2an-1an=2an-1故{an}是首项为a1=-1,公比为2的等差数列an=-2^(n-1)(2)设an=a+(n-1)d,bn=bq^(n-1)a5=a+4d=10因为{bn}是等比数列b3^2=b1*b5,所以a7^2=a5a10即(a+6d)^2=(a+4d)(a+9d)解方程a=0,d=5/2,a20=a+19d=95/2b1=b=a5=10,b3=bq^2=a7=15,b5=a10=45/2=bq^4解得b=10,q=√6/2或-√6/2,所以bn=10*(√6/2)^(n-1)或bn=10*(-√6/2)^(n-1)...
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