Tính tổng sau: A=11.2+12.3+13.4+...+199.100A = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}A=1.21+2.31+3.41+...+99.1001
A=1A = 1A=1
A=0A = 0A=0
A=12A = \frac{1}{2}A=21
A=99100A = \frac{{99}}{{100}}A=10099
Đáp án: <p><span data-type="math-inline" data-latex="A = \frac{{99}}{{100}}">$A = \frac{{99}}{{100}}$</span></p>
Phương pháp giải:
Sử dụng công thức 1n(n+1)=1n−1n+1\frac{1}{{n\left( {n + 1} \right)}} = \frac{1}{n} - \frac{1}{{n + 1}}n(n+1)1=n1−n+11
Lời giải chi tiết:
A=11.2+12.3+13.4+...+199.100=(1−12)+(12−13)+(13−14)+...+(199−1100)=1−12+12−13+13−14+...+199−1100=1−1100=99100\begin{array}{l}A = \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{99.100}}\\ = \left( {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 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{99}} - \frac{1}{{100}}\\ = 1 - \frac{1}{{100}} = \frac{{99}}{{100}}\end{array}A=1.21+2.31+3.41+...+99.1001=(1−21)+(21−31)+(31−41)+...+(991−1001)=1−21+21−31+31−41+...+991−1001=1−1001=10099
