n |
i=1 |
n?1 |
i=1 |
1 |
n |
當n=1時,a1=1成立,故an=
1 |
n |
(II)bn=n?2n
Sn=1?21+2?22+3?23+…+n?2n①
2Sn=1?22+2?23+3?24+…+(n-1)?2n+n?2n+1②
由①-②得,-Sn=21+22+23++2n-n?2n+1
=
2(1?2n) |
1?2 |
故Sn=(n-1)?2n+1+2
(III)證明:cn=
1 |
n2 |
令f(x)=2x-x2
f′(x)=2xln2-2x,又ln2>ln
e |
1 |
2 |
故f′′(x)=2x(ln2)2-2≥f′′(5)>0
故f′(x)在[5,+∞)上單調遞增,故f′(x)≥f′(5)>0
故f(x)在[5,+∞)上單調遞增,故f(x)≥f(5)=7>0
故當n>4時,2n>n2恒成立,即
1 |
2n |
1 |
n2 |
故
n |
i=1 |
n |
i=1 |
1 |
2i |
1 |
2n |
又
1 |
n2 |
1 |
n(n?1) |
1 |
n?1 |
1 |
n |
故
n |
i=1 |
1 |
2 |
1 |
2 |
1 |
3 |
1 |
n?1 |
1 |
n |
1 |
n |
綜上可得,1?
1 |
2n |
n |
i=1 |