Defining a is a rational number that is not 1, and we call 1 1 a the inverse of a.

This is cyclical, a1=1 3; a2=3/2;a3 = -2, a4 = 1 3, then a2009 is 2 3

The solution is calculated as: a1=1 3

a2=3/2

a3=-2a4=1/3

a5=3/2

a6=-2a7=1/3

a8 = 1 3 The four adjacent numbers are observed as a cycle of 2009 4 = 502....1.So a2009=1 3

Upstairs and downstairs have the answers... I don't need to go into details.

The solution is calculated as: a1=1 3

a2=3/2

a3=-2a4=1/3

a5=3/2

a6=-2a7=1/3

Through this set of calculations, we know that these numbers are three in a cycle, so we use 2009 3=502··· 1

The first one is 3 1 ... So...

a1=1/3;a2=3/2;a3=-2,a4=1/3,a5=3/2;a6=-2,a7=1/3,a8=3/2;a9=-2, three one cycle, to a2007 a total of 669 groups, then a2008 = 1 3, a2009 = 3 2.

There are also values to calculate a2008 a2009 a2010 and the result is -1.

It is known that a1 = 1 3 and a2 is the reciprocal difference of a1.

a2=1/(1-1/3)=3/2

a3 is the reciprocal difference of a2.

a3=1/(1-3/2)=-2

a4 is the inverse of a3, a4 = 1 (1 + 2) = 1 3 = a1

1 cycle every 3 rounds The cycle is 3

So a2011=a1=1 3,7, definition: a is a rational number that is not 1, and we call 1 1-a the reciprocal of the difference of a.

For example, the reciprocal difference of 2 is 1 1-2=-1, and the reciprocal difference of -1 is 1 1+1=1 2It is known that a1 = 1 3, a2 is the inverse difference of a1, a2 = ( a3 is the inversion difference of a2, a3 = ( a4 is the inverse difference of Li Hang A3, a4 = ( And so on, then a2011 = (

a1=1/3

a2=1/(1-a1)=3/2

a3=1/(1-a2)=-2

a4=1/(1-a3)=1/3

Because a4=a1

So it's 3 in a cycle.

2013 3 and 3 3 remainder are the same.

Therefore, a2013=a3=-2,1, definition: a is not a rational number for the skin to know 1, we call 1 1-a the reciprocal of the difference of holding a, for example, the reciprocal of 2 is 1 1-2=-1, and the reciprocal of the difference of -1 is 1 1-(-1)=1 2It is known that a1 = 1 3, a2 is the reciprocal of the difference of a1, and a3 is the reciprocal of the difference of a2. And so on, then positive elimination a2 = a3 = a4 = a2013 =

a1=1/3

a2=1/(1-1/3)=3/2

a3=1/(1-3/2)=-2

a4=1/[1-(-2)]=1/3

then a4=a1

So 3 in a cycle.

The 2013 3 remainder is the same as 3 3.

So a2013=a3=-2

From the title: a1 = 1 3

According to the definition, it can be solved: a2 = 1 (1-1 3) and knowing-blinding = 3 2;

a3=1 (1-3 mask2)=-2;

a4=1/（1-（-2））=1/3；

a5=3/2；

a6=-2；

See the pattern? Every 3 digits are combined with the state cycle once, 2012 3 = 670 more than 2, that is, a2012 = a (670 * 3 + 2) = a2 = 3 2

See the pattern as:

The reciprocal difference of a is 1 (1-a).

4、a1, a2, a3, a4, a5, a6, …1 3, 3 2, -2, 1 3, 3 2, -2 can be seen as period 3

2010 3=670 just happens to be a complete cycle, you can know that it is the last number of a cycle -2

Come to think of it, that's the key].

So a2010=-2

a1 is 1 3 a2 is 3 2 a3 is -2 a4 is 1 3 Gu is three one cycle a2010=a3 is -2

Test Center: Regularity: Variation of numbers

Topic: Regularity

Analysis: 11-a is called the reciprocal of the difference of a, knowing a1=-1 3, you can calculate a2, a3, a4, a5 in turn, you can find that every 3 numbers are a cycle, and then divide 2010 by 3 to get the answer

Answer: Solution: It is known that a1=-13, the reciprocal of the difference of a1 a2=11-(-13)=3 4;

The reciprocal of the difference of a2 a3=11 -3 4=4;

The reciprocal of the difference of a3 a4=11-4=-1 3;

The reciprocal of the difference of a4 a5 = 11 - (-13) = 3 4;

So, a2010=a3=4

So the answer is: 4

Comments: This question mainly tests students' understanding and mastery of reciprocal and numerical change knowledge points, and the key to solving this question is to calculate A2, A3, A4, and A5 in turn to find out the law of number change

Solution: complex a2=1 (1-a1)=1 [1-(-1 3)]=3 4a3=1 (1-a2)=1 (1-3 4)=4a4=1 (1-a3)=1 (1-4)=-1 3=a1 visible, every 3 operations, just.

The system changes bai into the original number, that is, duan is a series of numbers with a period of 3, and dao then a2016=a(3+3 671)=a3=4

Solution: a1=3

a2=1 (1-a1)=1 (1-3)=- a3=1 (1-a2)=1 [1-(-= a4=1 (1-a3)=1 (1- )=3=a1 The series starts from the plural 1, system.

Press 3,- cycles, one bai per 3 items.

2015 3=671 surplus du2, looped 671 times, zhi 672nd looped to dao2 item, is -

a2015=-½