1、 1/[n(n+ 1)]=( 1/n)-[ 1/(n+ 1)]

2、 1/[(2n- 1)(2n+ 1)]= 1/2[ 1/(2n- 1)- 1/(2n+ 1)]

3、 1/[n(n+ 1)(n+2)]= 1/2 { 1/[n(n+ 1)]- 1/[(n+ 1)(n+2)]}

4、 1/(√a+√b)=[ 1/(a-b)](√a-√b)

5、n n！ =(n+ 1)！ -No!

6、 1/[n(n+k)]= 1/k[ 1/n- 1/(n+k)]

7、 1/[√n+√( n+ 1)]= √( n+ 1)-√n

8、 1/(√n+√n+k)=( 1/k)[√( n+k)-√n]

Extended data:

Characteristics of elimination of division period

1. The positions before and after the other items are symmetrical.

2. The positivity and negativity of other projects are opposite.

Precautions for use

Pay attention to check whether the formula after splitting the item is equal to the original formula. The typical error is:1/(3× 5) =1/3-1/5 (the right side of the equation should be divided by 2).

Common methods for summation of series;

Formula method, split term elimination method, dislocation subtraction method, reverse order addition method, etc. (The key is to find the general item structure of the sequence)

1. Find the sum of series by grouping method: for example, an=2n+3n.

2. Sum by dislocation subtraction: for example, an = n 2n.

3. Sum by split term method: for example, an= 1/n(n+ 1)

4. sum up in reverse order: for example, an = n.