當(dāng)n≥2時(shí).Cn-Cn-1=230-100×1.05n-2 查看更多

 

題目列表(包括答案和解析)

對(duì)n∈N*,不等式
x>0
y>0
y≤-nx+2n
所表示的平面區(qū)域?yàn)镈n,把Dn內(nèi)的整點(diǎn)(橫坐標(biāo)與縱坐標(biāo)均為整數(shù)的點(diǎn))按其到原點(diǎn)的距離從近到遠(yuǎn)排成點(diǎn)列:(x1,y1),(x2,y2),(x3,y3),…,(xn,yn).
(1)求xn,yn;
(2)數(shù)列{an}滿足a1=x1且n≥2時(shí),an=yn(
1
2y1
+
1
2y2
+
1
2y3
+…+
1
2yn
)
,求數(shù)列{an}的前n項(xiàng)和Sn;
(3)設(shè)c1=1,當(dāng)n≥2時(shí),cn=lg[2
y
2
_
•(1-
1
y
2
2
)•(1-
1
y
2
3
)•(1-
1
y
2
4
)•…•(1-
1
y
2
n
)]
,且數(shù)列{cn}的前n項(xiàng)和Tn,求T99

查看答案和解析>>

設(shè)正整數(shù)數(shù)列{an}滿足a1=2,a2=6,當(dāng)n≥2時(shí),有|
a
2
n
-an-1an+1| <  
1
2
an-1

(1)求a3的值;(2)求數(shù)列{an}的通項(xiàng);
(3)記Tn=
12
a1
+
22
a2
+
32
a3
 +K+
n2
an
,證明:對(duì)任意n∈N*,Tn
9
4

查看答案和解析>>

已知數(shù)列{an}滿足a1=1,a2=3,且an+2=(1+2|cos
2
|)an+|sin
2
|,n∈N*
,
(1)求a2k-1(k∈N*);
(2)數(shù)列{yn},{bn}滿足y=a2n-1,b1=y1,且當(dāng)n≥2時(shí)bn
=y
2
n
(
1
y
2
1
+
1
y
2
2
+…+
1
y
2
n-1
)
.證明當(dāng)n≥2時(shí),
bn+1
(n+1)
-
bn
n2
=
1
n2
;
(3)在(2)的條件下,試比較(1+
1
b1
)•(1+
1
b2
)•(1+
1
b3
)+…+(1+
1
bn
)
與4的大小關(guān)系.

查看答案和解析>>

已知函數(shù)f(x)=
2x2+1
(x>0)
,數(shù)列{an}滿足a1=1,當(dāng)n≥2時(shí),an=f(an-1
(1)求an; 
(2)若bn=
2n
an+an+1
,若Sn=b1+b2+…+bn,求
lim
n→∞
bnSn
(an)2

查看答案和解析>>

(2013•成都一模)在數(shù)列{an}中,a1=2,a2=4,且當(dāng)n≥2時(shí),a
 
2
n
=an-1an+1
,n∈N*
(I)求數(shù)列{an}的通項(xiàng)公式an;
(II)若bn=(2n-1)an,求數(shù)列{bn}的前n項(xiàng)和Sn;
(III)求證:
1
a1
+
1
2a2
+
1
3a3
+…+
1
nan
3
4

查看答案和解析>>


同步練習(xí)冊(cè)答案