公式:

(1) 1n+a1n+b=1ba(1n+a1n+b)frac{1}{n+a} cdot frac{1}{n+b} = frac{1}{b-a} (frac{1}{n+a} - frac{1}{n+b})

(2) 1n(n+1)(n+k)=1k[1n(n+1)(n+k1)1(n+1)(n+2)(n+k)]frac{1}{n(n+1)…(n+k)} = frac{1}{k}[frac{1}{n(n+1)…(n+k-1)} - frac{1}{(n+1)(n+2)…(n+k)}]

其中: n=1,2,3… ; k表示有限项;

举例:

(1) 123+134++1(n+1)(n+2)frac{1}{2 * 3}+frac{1}{3 * 4} +…+ frac{1}{(n+1)(n+2)}

其中: 1(n+1)(n+2)=121[1(n+1)1(n+2)]frac{1}{(n+1)(n+2)} = frac{1}{2-1}[frac{1}{(n+1)}-frac{1}{(n+2)}]

=11(1213+1314++1(n+1)1(n+2))=frac{1}{1}(frac{1}{2} - frac{1}{3} + frac{1}{3} - frac{1}{4} +…+ frac{1}{(n+1)} - frac{1}{(n+2)})
=121(n+2)=frac{1}{2} - frac{1}{(n+2)}

(2) 1123+1234++1n(n+1)(n+2)frac{1}{1 * 2 * 3} + frac{1}{2 * 3 * 4} + … +frac{1}{n(n+1)(n+2)}

其中: 这里的k=2, 所以 1n(n+1)(n+2)=12[1n(n+1)1(n+1)(n+2)]frac{1}{n(n+1)(n+2)} = frac{1}{2}[frac{1}{n*(n+1)}-frac{1}{(n+1)(n+2)}]

=12(112123+123134++1n(n+1)1(n+1)(n+2))=frac{1}{2}(frac{1}{1 * 2} - frac{1}{2 * 3} + frac{1}{2 * 3} - frac{1}{3 * 4} +…+ frac{1}{n * (n+1)} - frac{1}{(n+1)(n+2)})
=12(121(n+1)(n+2))=frac{1}{2} * (frac{1}{2} - frac{1}{(n+1)(n+2)})

注意:本公式属于中学知识,必须掌握!