a) \(\displaystyle{\rm{}}{1 \over n}.{1 \over {n + 1}} = {1 \over {n(n + 1)}}\) \((1) \;\; (n ∈ Z, n ≠ 0)\)
\(\displaystyle{1 \over n} - {1 \over {n + 1}} = {1 \over n} + {{ - 1} \over {n + 1}} \)
\(\displaystyle= {{n + 1} \over {n(n + 1)}} + {{ - n} \over {n(n + 1)}} \)\(\displaystyle= {{n + 1 - n} \over {n(n + 1)}} \)
\(\displaystyle= {1 \over {n(n + 1)}}\) \((2)\)
Từ \((1)\) và \((2)\) ta có: \(\displaystyle{1 \over n}.{1 \over {n + 1}} = {1 \over n} - {1 \over {n + 1}}\)\(\left( {n \in Z,n > 0} \right).\)
b) Áp dụng kết quả câu a ta có:
\(\displaystyle{\rm{A}} = {1 \over 2}.{1 \over 3} + {1 \over 3}.{1 \over 4} + {1 \over 4}.{1 \over 5} + {1 \over 5}.{1 \over 6} \)\(\displaystyle + {1 \over 6}.{1 \over 7} + {1 \over 7}.{1 \over 8} + {1 \over 8}.{1 \over 9}\)
\(\displaystyle A= {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4} + {1 \over 4} - {1 \over 5} \)\(\displaystyle + {1 \over 5} - {1 \over 6} + {1 \over 6} - {1 \over 7} + {1 \over 7} - {1 \over 8}\)\(\displaystyle + {1 \over 8} - {1 \over 9} \)
\(\displaystyle A={1 \over 2} - {1 \over {9}} = {{9} \over {18}} - {{ 2} \over {18}} = {7 \over {18}} \)
\(\displaystyle B = {1 \over {30}} + {1 \over {42}} + {1 \over {56}} + {1 \over {72}} + {1 \over {90}} \)\(\displaystyle + {1 \over {110}} + {1 \over {132}}\)
\(\displaystyle = {1 \over 5}.{1 \over 6} + {1 \over 6}.{1 \over 7} + {1 \over 7}.{1 \over 8} + {1 \over 8}.{1 \over 9} \)\(\displaystyle + {1 \over 9}.{1 \over {10}} + {1 \over {10}}.{1 \over {11}} + {1 \over {11}}.{1 \over {12}}\)
\(\displaystyle = {1 \over 5} - {1 \over 6} + {1 \over 6} - {1 \over 7} + {1 \over 7} - {1 \over 8} \)\(\displaystyle + {1 \over 8} - {1 \over 9} + {1 \over 9} - {1 \over {10}} + {1 \over {10}} - {1 \over {11}} \)\(\displaystyle + {1 \over {11}} - {1 \over {12}} \)
\(\displaystyle = {1 \over 5} - {1 \over {12}} = {{12} \over {60}} - {{ 5} \over {60}} = {7 \over {60}} \)
