等比数列{an}中,a1+an=66 a2*An_1=128 sn=126 求n的公比q?注:
人气:457 ℃ 时间:2020-05-20 16:27:37
解答
A2*An-1=A1*An=128 A1+An=66 =>A1,An为方程xx-66x+128=0两根 =>A1,An=2,64或64,2 若A1=2,An=64 =>q^(n-1)=32 Sn=A1(1-q^n)/(1-q)=2*(1-32q)/1-q=126 => q=2,n=6 若A1=64,An=2 =>q^(n-1)=1/32 Sn=A1(1-q^n)/(1-q)=64*...
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