Tính các tổng saua) \(A=\frac{1}{2}+\frac{1}{6}+\f...

A

- Hướng dẫn giải

Phương pháp giải:

Áp dụng công thức \(\frac{1}{n.(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)

Giải chi tiết:

\(\begin{array}{l}a)A = \frac{1}{2} + \frac{1}{6} + \frac{1}{{12}} +  \ldots  + \frac{1}{{99.100}}\\ = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} +  \ldots  + \frac{1}{{99.100}}\\ = \left( {\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ... + \left( {\frac{1}{{99}} - \frac{1}{{100}}} \right)\\ = 1 + \left( {\frac{{ - 1}}{2} + \frac{1}{2}} \right) + \left( {\frac{{ - 1}}{3} + \frac{1}{3}} \right) + \left( {\frac{{ - 1}}{4} + \frac{1}{4}} \right) + ... + \left( {\frac{{ - 1}}{{99}} + \frac{1}{{99}}} \right) - \frac{1}{{100}}\\ = 1 + 0 + 0 + ... + 0 - \frac{1}{{100}}\\ = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}.\end{array}\)

\(\begin{array}{l}b)B = \frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ... + \frac{1}{{32.35}}\\ \Rightarrow 3.B = 3.\left( {\frac{1}{{2.5}} + \frac{1}{{5.8}} + \frac{1}{{8.11}} + ... + \frac{1}{{32.35}}} \right)\\3B = \frac{3}{{2.5}} + \frac{3}{{5.8}} + \frac{3}{{8.11}} + ... + \frac{3}{{32.35}}\end{array}\)

Ta có

\(\begin{array}{l}\frac{3}{{2.5}} = \frac{{5 - 2}}{{2.5}} = \frac{5}{{2.5}} - \frac{2}{{2.5}} = \frac{1}{2} - \frac{1}{5};\\\frac{3}{{5.8}} = \frac{{8 - 5}}{{5.8}} = \frac{8}{{5.8}} - \frac{5}{{5.8}} = \frac{1}{5} - \frac{1}{8};......\end{array}\)

Từ đó ta có:

\(\begin{array}{l}3B = \left( {\frac{1}{2} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{8}} \right) + \left( {\frac{1}{8} - \frac{1}{{11}}} \right) + ... + \left( {\frac{1}{{32}} - \frac{1}{{35}}} \right)\\3B = \frac{1}{2} + \left( {\frac{{ - 1}}{5} + \frac{1}{5}} \right)+\left( {\frac{{ - 1}}{8} + \frac{1}{8}} \right)+\left( {\frac{{ - 1}}{{11}} + \frac{1}{{11}}} \right) + ...+\left( {\frac{{ - 1}}{{32}} + \frac{1}{{32}}} \right) - \frac{1}{{35}}\\3B = \frac{1}{2} + 0 + 0 + ... + 0 - \frac{1}{{35}}\\3B = \frac{1}{2} - \frac{1}{{35}} = \frac{{33}}{{70}}\\ \Rightarrow B = \frac{{33}}{{70}}:3 \Rightarrow B = \frac{{11}}{{70}}\end{array}\)