题目
题型:不详难度:来源:
2 |
5 |
an |
an+1 |
4an+2 |
an+1+2 |
(1)求证:数列{
1 |
an |
(2)令bn=an•an+1,Tn=b1+b2+b3+…+bn,求证:Tn<
4 |
15 |
答案
an |
an+1 |
4an+2 |
an+1+2 |
∴2an-2an+1=3anan+1
两边同时除以anan+1可得,
1 |
an+1 |
1 |
an |
3 |
2 |
∴数列列{
1 |
an |
1 |
a1 |
5 |
2 |
3 |
2 |
∴
1 |
an |
5 |
2 |
3 |
2 |
3n+2 |
2 |
∴an=
2 |
3n+2 |
(2)bn=an•an+1=
2 |
3n+2 |
2 |
3n+5 |
4 |
3 |
1 |
3n+2 |
1 |
3n+5 |
∴Tn=b1+b2+b3+…+bn=
4 |
3 |
1 |
5 |
1 |
3n+5 |
4 |
15 |
核心考点
试题【已知数列{an}满足a 1=25,且对任意n∈N*,都有anan+1=4an+2an+1+2.(1)求证:数列{1an}为等差数列,并求{an}的通项公式;(2】;主要考察你对等差数列等知识点的理解。[详细]
举一反三
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设{bn}是等差数列,且b2=a2,b4=a4.求数列{bn}的公差,并计算b1-b2+b3-b4+______-b100的值.
(1)求a4及Sn;
(2)令bn=