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设{an}的公差为d，则a2=1+d,a3=1+2d,a4=1+3d{ bn }为等比数列,所以b3^2=b2·b4,即(a2+a4)^2=a3,(1+d+1+3d)=1+2d,解得d=-1/2或d=-3/8.b3=a2+a4=2+4d,当d=-1/2时b3=0,所以d=-1/2舍去当d=-3/8时b3=1/2,所以等比数列{bn}的公比q=√b3/b1=±√2/2所以Sn=na1+n(n-1)d/2=(-3n^2+19n)/16Tn=b1(1-q^n)/(1-q)=[1±(√2/2)^n]/(1±√2/2）希望对你有帮助... |
