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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)
a) =3/2 . 4/3 . 5/4 ...100/99
=\(\frac{3.4.5...100}{2.3.4..99}\)
=\(\frac{100}{2}\)
b) =
![](https://rs.olm.vn/images/avt/0.png?1311)
theo công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
=>\(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{2013}.\frac{2013.2014}{2}\)
\(=>A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2014}{2}=>A=\frac{1}{2}\left(1+2+3+..+2014\right)-\frac{1}{2}\)
\(=>A=\frac{1}{2}.\frac{2014.2015}{2}-\frac{1}{2}=1014552\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
A= \(\frac{100}{2}\)
A=50
2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)
B= \(\frac{1}{99}\)
\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)
\(=\frac{100}{2}\)
\(=50\)
\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)
\(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)
\(=-\frac{1}{99}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+\(\frac{1+2}{2}\)+\(\frac{1+2+3}{3}\)+...+\(\frac{1+2+3+...+16}{16}\)
A=\(\frac{2}{2}\)+\(\frac{3}{2}\)+\(\frac{4}{2}\)+...+\(\frac{17}{2}\)
A=\(\frac{2+3+4+...+17}{2}\)
A=76(đề thi HSG huyện tui có tui làm zậy mà cũng có điểm tuyệt đối)
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+....+\frac{1}{16}.\left(1+2+3+....+16\right)\)
\(A=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+\frac{1}{4}\cdot\frac{4.5}{2}+.....+\frac{1}{16}\cdot\frac{16.17}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}\)
\(A=\frac{\left(2+3+4+.....+17\right)}{2}=\frac{\left(2+17\right).16}{2}=\frac{152}{2}=76\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+...+100}\right)\)
\(A=\frac{2}{\left(1+2\right).2:2}.\frac{5}{\left(1+3\right).3:2}.\frac{9}{\left(1+4\right).4:2}...\frac{5049}{\left(1+100\right).100:2}\)
\(A=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{10098}{100.101}\)
\(A=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{99.102}{100.101}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{4.5.6...102}{3.4.5...101}\)
\(A=\frac{1}{100}.\frac{102}{3}=100.34=\frac{1}{100}.34=\frac{17}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
lam on ai biet thi chi trong toi nay tui se cho ma ngay mai la phai nop rui