还是数列!

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还是数列!

还是数列!
还是数列!
 

还是数列!
设 3个正数 分别为,a-c,a,a+c.则,
15 = (a-c)+a+(a+c) = 3a,a=5.
b(n) = bq^(n-1).
b(3) = a-c+2 = 7-c=bq^2.
b(4) = a+5=10=bq^3,
b(5) = a+c+13=18+c = bq^4,
10^2 = [b(4)]^2 = b(3)b(5) = (7-c)(18+c) = 126 - 11c - c^2,
0 = c^2 + 11c - 26 = (c+13)(c-2),
c=-13或c=2.
若c=-13,则a+c=5-13=-8与a+c>0矛盾.
因此,只能 c=2.
此时,
b(3) = 7-c = 5 = bq^2,
b(4) = 10 = bq^3.q = bq^3/(bq^2) = 2.
b(5) = 18+c=20 = bq^4 = b*16,b = 5/4.
b(n) = (5/4)2^(n-1) = 5*2^(n-3).
s(n) = (5/4)[2^n - 1]/(2-1) = (5/4)[2^n - 1]= (5/2)2^(n-1) - 5/4,
s(n) + 5/4 = (5/2)2^(n-1),
{s(n)+5/4}是首项为5/2,公比为2的等比数列.