等比数列的填空题求解，要详细。 求解，两道等比数列数学选择题
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\u5e0c\u671b\u5bf9\u4f60\u6709\u5e2e\u52a9\u8bf7\u91c7\u7eb3
a<n+1> = 5S<n> + 1 = 5a1(q^n-1)/(q-1) + 1 (1)
a<n> = 5S<n-1> + 1 = 5a1[q^(n-1)-1]/(q-1) + 1 (2)
两式相除 (1)/(2) 得
q = [5a1(q^n-1)/(q-1) + 1] / {5a1[q^(n-1)-1]/(q-1) + 1}
= [5a1(q^n-1) + q-1] / {5a1[q^(n-1)-1] + q-1}
5a1(q^n-1) + q-1 = q {5a1[q^(n-1)-1] + q-1}
5a1q^n-5a1 + q-1 = 5a1q^n-5a1q + q^2-q
5a1(q-1) = q^2-2q+1 = (q-1)^2
5a1 = q-1 (3)
由（1） a<2> = 5S<1> + 1 = 5a1+ 1,
即 a1q = 5a1+1, a1(q-5) = 1 （4）
（3）（4）联立解得 q = 6, q = 0(舍弃)
an+1=Sn+1-Sn=5Sn+1，故Sn+1=6Sn+1，显然q≠1，而Sn=a1（1-q^n）/（1-q），带入上式可得q=6
由题意得，
为等比数列
综上所述，答案为：
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