一点都不难.说困扰已久,假的吧.
1.
a(n)=a(1) x q^(n-1)
a(1) X a(3) = (a(1) x q)^2 = 4 => a(1) X q = 2
S3 = a(1) + a(2) + a(3) = a(1) X (1 + q + q^2) = 2 X (1 + q + q^2) / q = 7
2 q^2 - 5 q + 2 = 0
q = 1/2 或 2,
q > 1
因此 q = 2,a(1) = 1,a(5) = 2^4 = 16
2.
b(n) = 3 X 2^n - 1
Sn = 3 X 2(2^n - 1) - n = 6 X 2^n -n - 6
3.
a(2) + a(4) + ...+ a(100) = a(1) + a(3) + ...+ a(99) + 99 * d = 150 + 99 = 249