An为等比数列,且Am+1加Am-1=Am的平方,已知前S2m-1=38,m=?凌乱了,

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a(n) = aq^(n-1),
[a(m)]^2 = a^2q^(2m-2) = a(m+1) + a(m-1) = aq^m + aq^(m-2),
0 = aq^(m-2)[aq^m - q^2 - 1],
0 = aq^m - q^2 - 1.
若q=1,则s(n) = na,38 = s(2m-1) = (2m-1)a,
0 = aq^m - q^2 - 1 = a - 2,a = 2,
38 = 2(2m-1),
19 = 2m-1,
m=10.
若q不为1,则s(n) = a[1-q^n]/(1-q).
0 = aq^m - q^2 - 1,
a = (q^2 + 1)/q^m.
38 = s(2m-1) = a[1-q^(2m-1)]/(1-q) = (q^2+1)[1-q^(2m-1)]/[(1-q)q^m].
0 = 38(1-q)q^m - (q^2 + 1)[1-q^(2m-1)] = 38q^m - 38q^(m+1) - q^2 - 1 + q^(2m+1) + q^(2m-1),
一个方程,包含2个变量(q,m),无法解出.
综合有,q=1,a=2,a(n) = 2,s(n) = 2n,m=10满足题意.

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