\(E=\frac{30}{1.2.3}+\frac{30}{2.3.4}+\frac{30}{3.4.5}+...+\frac{30}{98.99.100}\)

\(E=15\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(E=15\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(E=15\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(E=15.\frac{4949}{9900}\)

\(E=\frac{4949}{660}\)

+ \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\cdot\frac{2}{n\left(n+1\right)\left(n+2\right)}\) \(=\frac{1}{2}\cdot\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)

Do đó : \(E=30\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\right)\)

\(E=30\cdot\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(E=15\cdot\left(\frac{1}{2}-\frac{1}{9900}\right)=15\cdot\frac{4949}{9900}=\frac{4949}{660}\)

1) Đặt \(A=1.2+2.3+3.4+....+98.99\)

Ta có:\(3A=3.\left(1.2+2.3+3.4+....+98.99\right)\)

\(3A=1.2.3+2.3.3+3.4.3+....+98.99.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)

\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}=323400\)

Ta có:\(\frac{A.y}{1}=184800\Rightarrow y=184800:323400=\frac{4}{7}\)

2)Đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\right).1428+185,8\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{37.38.39}\)

Tổng quát:\(\frac{2}{\left(a-1\right)a\left(a+1\right)}=\frac{1}{\left(a-1\right)a}-\frac{1}{a\left(a+1\right)}\)

Ta có:

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+.....+\frac{2}{37.38.39}\)

\(2B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+\left(\frac{1}{3.4}-\frac{1}{4.5}\right)+...+\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(2B=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\Rightarrow B=\frac{370}{741}:2=\frac{185}{741}\)

Khi đó \(A=\frac{185}{741}.1428+185,8=...........\) (tự tính ra)

(*)số ko đẹp mấy

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(A=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

chỗ nãy rồi bạn tự tính tiếp

KQ la \(\frac{4949}{19800}\)ak cac ban

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{4949}{19800}\)

A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)

=1/2*(1/2-1/99*100)

=1/2*(4950-1/9900)

=4950/19800

\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)

\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)

\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)

A=1/2.(2/1.2.3+2/2.3.4+...+2/98.99.100

=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)

\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)

\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)

\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)

\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)