等比數(shù)列{An}的前n項(xiàng)和為Sn,已知對(duì)任意的n∈N+點(diǎn)（n,Sn）均在函數(shù)y+b^x+r(b>0)且b≠1,b,r均為常數(shù)）圖像上

等比數(shù)列{An}的前n項(xiàng)和為Sn,已知對(duì)任意的n∈N+點(diǎn)（n,Sn）均在函數(shù)y+b^x+r(b>0)且b≠1,b,r均為常數(shù)）圖像上
(1)求r的值
（2）當(dāng)b=2時(shí),記Bn=（n+1）/4An(n∈N+)求數(shù)列{Bn}的前n項(xiàng)的Tn.

數(shù)學(xué)人氣：434 ℃時(shí)間：2020-04-16 06:48:26

優(yōu)質(zhì)解答

由題意可知,Sn=b^n+r
所以
An = Sn - S = b^n - b^(n-1)
A = b^(n-1) - b^(n-2)
An/A = b
所以An數(shù)列的公比為 b
則
Sn = A1 * (b^n -1)/(b-1) = [A1/(b-1)]*b^n - [A1/(b-1)]
同時(shí)
Sn = b^n + r
若對(duì)任意n, 以上2式子同時(shí)成立, 則
A1/(b-1) = 1
r = -1
---------------------------
當(dāng) b = 2 時(shí)
A1 = 1
An = A1 * b^(n-1) = 2^(n-1)
Sn = A1 * (b^n -1)/(b-1) = 2^n -1
Bn = n + 1/(4An)
= n + 1/2^(n+1)
Tn = B1 + B2 + …… + Bn
= (1 + 2 + …… + n) + [1/2^2 + 1/2^3 + …… + 1/2^(n+1)]
= n(n+1)/2 + (1/4)*[(1/2)^n - 1]/(1/2 - 1)
= n(n+1)/2 - (1/2)*[(1/2)^n -1]

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等比數(shù)列{An}的前n項(xiàng)和為Sn,已知對(duì)任意的n∈N+點(diǎn)（n,Sn）均在函數(shù)y+b^x+r(b>0)且b≠1,b,r均為常數(shù)）圖像上

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