Question

设数列an为等差数列,数列bn为等比数列若a1<a2,b1<b2,bi=ai^2

设（an）为等差数列，（bn）为等比数列，若a1小于a2，b1小于b2，且bi等于ai的平方，（i等于1.2.3）,则数列bn的公比

Answer 1

a(n) = a + (n-1)d,
a = a(1) < a(2) = a + d, d>0.

b(n) = bq^(n-1) = [a(n)]^2 >=0.
b = b(1) < b(2) = bq, b>0, q>1.
b = b(1) = [a(1)]^2 = a^2,

b(n) = a^2q^(n-1).

b(2) = a^2q = [a(2)]^2 = [a+d]^2, q = [1 + d/贺稿a]^2.

b(n) = a^2[1+d/a]^(2n-2)

b(3) = a^2[1+d/a]^4 = [a(3)]^2 = [a+2d]^2,

[1+d/a]^4 = [1+2d/a]^2,

0 = [(1+d/a)^2 - (1+2d/a)][(1+d/a)^2 + (1+2d/a)]
= [1 + 2d/a + (d/a)^2 - 1 - 2d/a][1+2d/a + (d/a)^2 + 1 + 2d/a]
= (d/a)^2[(d/a)^2 +4d/a + 2],

0 = (d/a)^2 + 4(d/a) + 4 - 2 = [d/a+2]^2 - 2 = [d/a+2+2^(1/2)][d/a+2-2^(1/2)],

0 = d/a + 2 + 2^(1/2)或0 = d/a + 2-2^(1/2).

q = [1+d/a]^2 = [1+2^(1/2)]^2 = 3 + 2^(3/2),
或
q = [1+d/a]^2 = [1-2^(1/2)]^2 = 3 - 2^(3/2)<袜裤1(舍告拍简去)

因此，只能，
q = 3 + 2^(3/2).

Answer 2