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11 tháng 4 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 10 2016
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{99.100}=\frac{49}{99.100}\Rightarrow A=\frac{49}{2.99.100}\)
TM
26 tháng 3 2017
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
15 tháng 2 2016
C = 1.2.3 + 2.3.3 + 3.3.4 + .... + 3.99.100
Đặt M = 1.2.3 + 2.3.4 + 3.4.5 + .... + 99.100.101
=> M - 3A = 1.2.3 - 1.2.3 + 2.3.(4-3) + 3.4 ( 5-3) + .... + 99.100 ( 101 -3)
= 1.2.3 + 2.3.4 + .... + 98.99.100
=> M -3A = M - 99.100.101
=> A = 99.100.101/3 = 333300
D =
Đặt A=1.3.5 + 3.5.7 + 5.7.9 + ................ + 95.97.99
8A= 1.3.5.8 + 3.5.7.8 + 5.7.9.8 + ................ + 95.97.99.8
8A=1.3.5(7+1)+3.5.7(9-1)+5.7.9.(11-3)+.......+95.97.99(101-93)
8A=3.5.7+15+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+.......+95.97.99.101-93.95.97.99
8A=15+95.97.99.101
A= \(\frac{15+95.97.99.101}{8}\)
A=11517600
NT
1
MS
23 tháng 4 2018
\(A=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=2.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=2\cdot\frac{4949}{9900}=\frac{4949}{4950}\)
MN
5 tháng 2 2016
\(A=\frac{11}{1.2.3}+\frac{11}{2.3.4}+\frac{11}{3.4.5}+...+\frac{11}{98.99.100}\)
\(A=\frac{11}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{11}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{11}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{11}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)=\frac{11}{2}.\left(\frac{4950}{9900}-\frac{1}{9900}\right)=\frac{11}{2}.\frac{4949}{9900}=\frac{4949}{1800}\)
DD
30 tháng 7 2015
A=1/1.2.3+1/2.3.4+...+1/98.99.100
2A=2/1.2.3+2/2.3.4+...+2/98.99.100
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=1/2-1/9900
2A=4949/9900
A=4949/19800
5 tháng 4 2018
A=1/1.2.3+1/2.3.4+...+1/98.99.100
2A=2/1.2.3+2/2.3.4+...+2/98.99.100
2A=1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=1/2-1/9900
2A=4949/9900
A=4949/19800
KH
12 tháng 5 2017
* Chứng tỏ
Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
\(4C=98\cdot99\cdot100\cdot101\)
\(C=98\cdot99\cdot25\cdot101=24497550\)