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a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)

\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+\frac{20}{5.7.9}+...+\frac{20}{25.27.29}\)
\(=5.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(=5.\frac{260}{783}\)
\(=\frac{1300}{783}\)
Ta có:\(\frac{1}{\left(n-2\right)n}-\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)-\left(n-2\right)n}{\left(n-2\right)n\cdot n\left(n+2\right)}\)
\(=\frac{n\left(n+2-n+2\right)}{n\cdot\left(n-2\right)n\left(n+2\right)}=\frac{4}{\left(n-2\right)n\left(n+2\right)}\)
Áp dụng\(\frac{20}{1.3.5}+\frac{20}{3.5.7}+...+\frac{20}{25.27.29}\)
\(=5\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(=5\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(=5\cdot\frac{261-1}{783}=5\cdot\frac{260}{783}=\frac{1300}{783}\)

minh chi can ket qua thoi cung duoc ko can giai ra dau
ai lam dung minh kick
Đặt biểu thức trên = A
Xét : B = 1.2.3+2.3.4+....+n.(n+1).(n+2)
4B = 1.2.3.4+2.3.4.4+....+n.(n+1).(n+2).4
= 1.2.3.4+2.3.4.(5-1)+....+n.(n+1).(n+2).[(n+3)-(n-1)]
= 1.2.3.4+2.3.4.5-1.2.3.4+....+n.(n+1).(n+2).(n+3)-(n-1).n.(n+1).(n+2)
= n.(n+1).(n+2).(n+3)
=> B = n.(n+1).(n+2).(n+3)/4
=> A = 222315.222316.222317.222318/4
k mk nha

___Vương Tuấn Khải___
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070
Theo bài ra ta có:
B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
Ta nhân cả 2 vế với số 4 thì được phương trình như sau;
4*B = 4*(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
<=> 4*B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
<=> 4*B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
<=> 4*B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
<=> 4*B = 17.18.19.20
<=> 4*B = 116280
<=> B = 116280/4 = 29070

\(B=\frac{1}{1.1.3}+\frac{1}{2.3.5}+\frac{1}{3.5.7}+\frac{1}{4.7.9}+...+\frac{1}{100.199.201}\)
< \(\frac{1}{1.1.3}+\frac{2}{2.3.5}+\frac{3}{3.5.7}+\frac{4}{4.7.9}+...+\frac{100}{100.199.201}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{199.201}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{199.201}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{199}-\frac{1}{201}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{201}\right)=\frac{1}{2}.\frac{200}{201}=\frac{100}{201}< \frac{1}{2}< \frac{2}{3}\)
=> B < 2/3

Bùi Lê Anh Khoa
Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070
Ta có: B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19
=> 4B = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 17.18.19)
=> 4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +17.18.19.4
=> 4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +17.18.19.(20 - 16)
=> 4B = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 16.17.18.19
=> 4B = 17.18.19.20
=> 4B = 116280
=> B = 29070

4N = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 2015.2016.2017.(2018-2014)
4N = 1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 2015.2016.2017.2018 - 2014.2015.2016.2017
4N = (1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ... + 2015.2016.2017.2018) - (0.1.2.3 + 1.2.3.4 + 2.3.4.5 + ... + 2014.2015.2016.2017)
4N = 2015.2016.2017.2018 - 0.1.2.3
4N = 2015.2016.2017.2018
N = 2015.2016.504.2018 (kq hơi to nên bn tự tính nhé)
Giúp mình vs mình đang cần gấp