Solution, junior high school mathematics, Xiao Shengchu said that he didn t know how to solve it

After learning about high school splits, this topic is relatively simple.

Now that the general term formula has been written, the essence is to change this general term into the subtraction form of two fractions, and the simplified operation can be carried out:

1/[（3n-2）（3n+1）]=a/（3n-2）-b/（3n+1）=（3na+a-3nb+2b）/[（3n-2）（3n+1）]=[（3a-3b）n+（a+2b）]/[（3n-2）（3n+1）]。

Comparing the before and after expressions of this equation, we get:

3a-3b=0，a+2b=1。Solve the system of equations: a=b=1 3.

So: 1 [(3n-2)(3n+1)]=(1 3)[1 (3n-2)-1 (3n+1)].

Original = (1 3) [1-1 4+1 4-1 7+1 7-1 10+1 10-1 13+.1/（3n-2）-1/（3n+1）]=（1/3）×[1-1/（3n+1）]=（1/3）×（3n）/（3n+1）=n/（3n+1）。

solution, analysis, 1 (3n-2)(3n+1 ) =1 3(1 (3n-2)-1 (3n+1)) = original formula = 1 3[1-1 4+1 4+,,1 (3n-2) -1 (3n+1)] =1 3x(3n (3n+1)) 2n (3n+1) i 氵三氵三 一丿

Let's look at the denominator first, there is 1 for 2 and 2 for 3.

Then you can find out what is the denominator of the 2002nd fraction.

The denominator can be found to be 63+1=64

2002-1953=49 (terms) can also be obtained

The denominator is 49 and the 2002nd fraction is 49 64

Solution: Looking at the denominator, the law is:

The denominator is 2, 1.

The denominator is 3 2.

The denominator is 4 3.

The denominator is n n-1).

Therefore, the above question can be converted to start from 1, and when to add, the closest to 2002

That is, to find (n+1)n 2 at what value n is close to 2002, that is, (n+1)n is close to 4004 when n is what value

Since the square of 4004 is about equal to 63, 63*64 4032 is closer to 4004, and the difference is 24<64, so it can be seen that the denominator of the 2002nd position is 64

Look at the numerator again: when n=63, there are (63-1)+1 *62 2 1953 terms, and there are 49 items left from 2002 terms, so the numerator is 49

Answer: So, item 2002 is 49 64

This should be a high school topic.

Calculated using the equation of equal difference.

For example, the denominator is 1 corresponding to 0 numbers, the denominator 2 corresponds to 1 number, the denominator 3 corresponds to 2 numbers, and the denominator 4 corresponds to 3 numbers... The corresponding numbers are 0,1,2,3... n into a series of equal differences and the denominator = number n-1

So 2002=0+1+2+3+. n

Use the equation of equal difference to calculate n, the molecule n-1

Note that if n is not an integer, first look at the number of the integer part, if it is x, then the denominator of the 2002nd is x+1.