Solution: an =1/[n (n+1)] = (1/n)-[1/(n+1)] (crack term)
Then sn =1-(1/2)+(1/2)-(1/3)+(1/4) ...+.
= 1- 1/(n+ 1)
= n/(n+ 1)
Example 2 Find the sum of the first n terms of the sequence an=n(n+ 1) by using the basic form of integer split terms.
Solution: an = n (n+1) = [n (n+1) (n+2)-(n-1) n (n+1)]/3 (split term)
Then sn = [1× 2× 3-0×/kloc-0 /× 2+2× 3× 4-1× 2× 3+...+n (n+1) (n+2)-(n-)
= [n(n+ 1)(n+2)]/3
Example 31/(/kloc-0 /× 4)+1/(4× 7)+1/(7×10)+...+1(9/kloc)
The original formula =1/3 * [(1-kloc-0//4)+(1/4-1/7)+(1/7).
This deformation is characterized by the fact that after each item in the original series is split into two items, most of the items in the middle cancel each other out. There are only a few things left.
Note: The remaining projects have the following characteristics.
1 The position of the remaining items is symmetrical before and after.
The positive and negative of the other items are opposite.
Error-prone point: Pay attention to check whether the formula after the split 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).
Attachment: 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.
5, the method of finding the maximum and minimum terms of the sequence:
① an+ 1-an = ... For example, an= -2n2+29n-3.
② (An>0) as a =
③ an=f(n) Study the increase and decrease of function f(n), such as an = an 2+bn+c (a ≠ 0).
6. In arithmetic progression, the problem of the maximum value of Sn is usually solved by the adjacent term sign change method;
(1) When A 1 >: 0, d<0, the number m of terms satisfying {an} makes Sm take the maximum value.
(2) when a 1
7. The formula1/n+1/(n+1)+1/(n+2) ...+1/(n+n) is also applicable.