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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=4.\frac{3200}{9603}=\frac{12800}{9603}\)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\frac{3200}{9603}\)
\(A=\frac{800}{9603}\)
Bài trc mik làm lộn :)))
~ Hok tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A=1.2.3+2.3.4+...+98.99.100
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+98.99.100.4
=1.2.3.4+2.3.4.(5-1) + 3.4.5.(6-2) +...+ 98.99.100.(101-97)
=1.2.3.4+2.3.4.5 -1.2.3.4+3.4.5.6 -2.3.4.5+...+98.99.100.101-97.98.99.100
= 98.99.100.101
Suy ra A=98.99.100.101
=24497550
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(C=1.2.3+2.3.4+3.4.5+...+99.100.101\)
\(4C=1.2.3.4+2.3.4.4+3.4.5.4+...+99.100.101.4\)
\(4C=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+....+99.100.101.\left(102-98\right)\)
\(4C=1.2.3.4+2.3.4.5+3.4.5.6+...+99.100.101.102\)
\(4C=99.100.101.102=101989800\)
\(\Rightarrow C=\frac{101989800}{4}=25497450\)
A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100).4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
A=98.99.100.101/4
![](https://rs.olm.vn/images/avt/0.png?1311)
$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$
![](https://rs.olm.vn/images/avt/0.png?1311)
1.3.5.8 + 3.5.7.8 + 5.7.9.8 + … + 95.97.99.8
= 1.3.5(7 + 1) + 3.5.7(9 - 1) + 5.7.9(11 - 3) + … + 95.97.99(101 - 93)
= 1.3.5.7 + 15 + 3.5.7.9 - 1.3.5.7 + 5.7.9.11 - 3.5.7.9 + … + 95.97.99.101 - 93.95.97.99
= 15 + 95.97.99.101
=> \(A=\frac{15.95+97.99.101}{8}\)
\(A=1.2.3+2.3.4+3.4.5+...+99.100.101\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+99.100.101.4\)
\(\Rightarrow4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+99.100.101.\left(102-98\right)\)
\(\Rightarrow4A=\left(1.2.3.4+2.3.4.5+...+99.100+101.102\right)-\left(0.1.2.3+1.2.3.4+...+98.99.100.101\right)\)
\(\Rightarrow4A=99.100.101.102-0.1.2.3\)
\(\Rightarrow A=\dfrac{99.100.101.102-0.1.2.3}{4}\)
\(B=1.3.5+3.5.7+5.7.9+...+95.97.99\)
\(\Rightarrow8B=1.3.5.8+3.5.7.8+5.7.9.8+...+95.97.99.8\)
\(\Rightarrow8B=1.3.5.\left(7+1\right)+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+...+95.97.99.\left(101-93\right)\)
\(\Rightarrow8B=1.3.5.7+1.1.3.5+3.5.7.9-1.3.5.7+5.7.9.11-3.5.7.9+...+95.97.99.101-93.95.97.99\)
\(\Rightarrow8B=1.1.3.5+95.97.99.101\)
\(\Rightarrow B=\dfrac{1.1.3.5+95.97.99.101}{8}\)
B= 912720