解答:解:(1)解法1:∵a
1+a
2=40,a
1a
2=256,且q>1,解得
---------------(2分)
∴
q==4,∴
an=a1qn-1=8×4n-1=22n+1---------------------------------(4分)
∴b
n=log
2a
n=
log222n+1=2n+1--------------------------------------------(6分)
解法2:由a
1+a
2=40,a
1a
2=256,且q>1得
,∴
q==4------------------------------------(2分)
∴
bn+1-bn=log2an+1-log2an=log=log24=2,----------------------------(3分)
又b
1=log
2a
1=log
28=3,-------------------------------------------------------(4分)
∴{b
n}是以3為首項(xiàng),2為公差的等差數(shù)列,----------------------------------------(5分)
∴b
n=3+(n-1)×2=2n+1;----------------------------------------------------(6分)】
(2)當(dāng)n≥2時(shí),T
n-T
n-1=b
n-1=2n-1,
∴T
n=(T
n-T
n-1)+(T
n-1-T
n-2)+…(T
3-T
2)+(T
2-T
1)+T
1=
(2n-1)+(2n-3)+…+5+3==(n-1)(n+1);---------------(8分)
∵當(dāng)n≥2時(shí),
==(-),----------------------------(10分)
∴
n |
|
i=2 |
=
+++…+=
[(1-)+(-)+(-)+…(-)+(-)+(-)]=
(1+--)=-(+).------(12分)
∵n≥2,∴
+≤+=∴
-(+)≥-•=.
又
+>0∴
-(+)<即對(duì)?n∈N
*,n≥2,
≤n |
|
i=2 |
<.----------------------------------------------(14分)