Question

已知数列｛an｝的前n项和为Sn，点（an+2,Sn+1

已知数列｛an｝的前n项和为Sn，点（an+2,S(n+1))在直线y=4x-5上。令bn=an+1-2an，且a1=1,若f(x)=b1x+b2x^2+b3x^3+.........+bnx^n,比较f‘(1)与8n^2-4n的大小
是 a(n) +2 不是a(n+2)
后面的an+1是a(n+1)

Answer 1

an+2 = a(n) +2 还是 a(n+2)?
后面的an+1呢？

假设是a(n+1)
解：
点（an+2,S(n+1))在直线y=4x-5上
得
s(n+1) = 4an +3 -----1
s(n+2) = 4a(n+1) +3 ----2
2式减去1式得
s(n+2) - s(n+1) = 4(a+1) - 4an
a(n+2) = 4(a+1) - 4an
a(n+2) - 2a(n+1) = 2(a(n+1) - 2an)
bn = a(n+1)-2an
b(n+1) = 2bn
即{bn}为等比数列
由s2=4a1+3=a1+a2且a1=1得
a2=6 b1=a2-2a1=4
所以bn=4.2^(n - 1)
f‘(1)=b1+2b2+......+nbn -----3
2f‘(1) = b2 + 2b3 + ......+nb(n+1) ----4
4式减去3式得
f‘(1) = nb(n+1) -(b1+b2+b3+......+bn)=4.n.2^n-4.2^n+4
令G(x)=4.x.2^x-4.2^x+4
K(x)=8x^2-4x
H(x) = G(x) - K(x) (x>=3)
H'(x) = 4.2^x - 4.ln2.2^x-4ln2.2^x - 4ln2.2^x - 16x + 4
=4(1-ln2)2^x + 16x(ln2.2^(x-2) -1) + 4 因为x>=3
所以H'(x)>=4ln(e/2).2^x + 16x(2ln2 - 1) + 4
=4ln(e/2).2^x + 16x.ln(4/e) + 4
ln(e/2) > ln(1)=0 ln(4/e)>ln(1) = 0
所以H'(x)>0即H(x) 在x>=3上递增
又H(3) = 8 所以当x>=3时有H(x) > 0
<=>G(x) > K(x)
所以当n>=3时 f'(1) > 8n^2-4n
n=1时 f'(1) = 4 8n^2-4n = 4 即 f'(1) = 8n^2-4n
n=2时 f'(1) = 20 8n^2-4n = 24 即 f'(1) < 8n^2-4n

综上所述：
n=1时 f'(1) = 8n^2-4n
n=2时 f'(1) < 8n^2-4n
n>=3时 f'(1) > 8n^2-4n