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

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

1/(√a+√b)=[ 1/(a-b)](√a-√b)

No, no! =(n+ 1)！ -No! ?

Extended data:

Example 1 Find the sum of the first n terms of the sequence an= 1/n(n+ 1) by using the basic form of fractional split terms.

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