Tính tổng:
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
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G6
3
KT
19 tháng 8 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{19.20.21}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{20.21}\right)\)
\(=\frac{1}{2}.\frac{209}{420}\)
\(=\frac{209}{840}\)
U
19 tháng 8 2018
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot21}\right)\)
bn tự lm tp
26 tháng 8 2018
Đặt C =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)
2 tháng 4 2019
Giải :
\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)
2 tháng 4 2019
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Study well ! >_<
24 tháng 4 2016
Đặt
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{37.38.39}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}=\frac{1}{1.2}-\frac{1}{38.39}=\frac{1}{2}-\frac{1}{1428}\Rightarrow A=\left(\frac{1}{2}-\frac{1}{1428}\right):2=\frac{713}{1428}.\frac{1}{2}\)
=>S=\(\frac{713}{1428}.\frac{1}{2}.1428+185.8=\frac{713}{2}+185.8=\frac{713}{2}+1480=356+\frac{1}{2}+1480=1836\frac{1}{2}\)
24 tháng 4 2016
S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)+..............+\(\frac{1}{2}\left(\frac{1}{37.38}-\frac{1}{38.39}\right)\).1428+185.8
S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+........\frac{1}{37.38}-\frac{1}{38.39}\right).\)1428+185.8
S=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\).1428+185.8
S=\(\frac{185}{741}\).1428+1480
S=1836,518219
NC
2
22 tháng 3 2017
Ta có công thức:
\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)
vậy
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)
mình làm đầu tiên đó,
Chúc bạn học tốt !
S
22 tháng 3 2017
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)
\(=\frac{1}{2}\cdot\frac{77}{156}\)
\(=\frac{77}{312}\)
TA
21 tháng 3 2015
T/c:A=1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+...+1/97*98*99+1/98*99*100
2A=2/1*2*3+2/2*3*4+2/3*4*5+2/4*5*6+...+2/97*98*99+1/98*99*100
2A=(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+.....+(1/97*98-1/98*99)+(1/98*99-1/99*100)
2A=1/2+1/99*100
A=tự tính nha
VA
1
11 tháng 4 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
PH
3
15 tháng 5 2019
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
BD
15 tháng 5 2019
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(C=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{999\cdot1000}\right]\)
Tính nốt :v
Ta có
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\Rightarrow2C=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{2}-\frac{1}{999000}\)
\(=\frac{499500}{999000}-\frac{1}{999000}\)
\(=\frac{499499}{999000}\)
\(\Rightarrow C=\frac{499499}{1998000}\)
đúng nha bạn nhớ k mik