Tính nhanh (nếu có thể):
a, \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^8}\)
b, \(\frac{-1}{100.99}+\frac{-1}{99.98}+\frac{-1}{98.97}+..........+\frac{-1}{3.2}+\frac{-1}{2.1}\)
K
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HT
13 tháng 7 2016
\(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-1+\frac{1}{100}\)
\(C=\frac{-49}{50}\)
13 tháng 7 2016
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - ... - 1/3.2 - 1/2.1
C = 1/100 - (1/100.99 + 1/99.98 + 1/98.97 + ... + 1/3.2 + 1/2.1)
C = 1/100 - (1/1.2 + 1/2.3 + ... + 1/98.99 + 1/99.100)
C = 1/100 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/98 - 1/99 + 1/99 - 1/100)
C = 1/100 - (1 - 1/100)
C = 1/100 - 99/100
C = -98/100 = -49/50
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
23 tháng 6 2016
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
Đặt biểu thức trong ngoặc đơn là B ta có
\(B=\frac{100-99}{100.99}+\frac{99-98}{99.98}+\frac{98-97}{98.97}+...+\frac{3-2}{3.2}+\frac{2-1}{2.1}\)
\(B=\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+\frac{1}{2}-\frac{1}{3}+1-\frac{1}{2}\)
\(B=1-\frac{1}{100}=\frac{99}{100}\)
\(C=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}\)
T
26 tháng 5 2019
#)Giải :
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+... +\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\left(1-\frac{1}{100}\right)\)
\(=-\frac{99}{100}\)
#~Will~be~Pens~#
26 tháng 5 2019
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right]\)
\(=-\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right]\)
\(=-\left[1-\frac{1}{100}\right]=-\frac{99}{100}\)
31 tháng 8 2016
C = 1/100 - ( 1/2.1 + 1/3.2 + ... + 1/98.97 + 1/99.98 + 1/100.99
C = 1/100 - ( 1- 1/2+ 1/2 - 1/3 + ... + 1/97 - 1/98 + 1/98 - 1/99 + 1/99 - 1/100 )
C = 1/100 - ( 1 - 1/100 )
C = 1/100 - 99/100
C = \(\frac{-49}{50}\)
HT
26 tháng 8 2015
C = \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-.....-\frac{1}{2.1}\)
C = \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
C = \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
C = \(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}\)
C = \(\frac{-49}{50}\)
LK
6 tháng 7 2019
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}.\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}=...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right).\)
\(=-\left(1-\frac{1}{100}\right)=-\frac{99}{100}\)
chúc bạn học tốt
8 tháng 9 2016
Ta có:\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{9900}-\left(\frac{1}{99.98}+\frac{1}{98.97}+\frac{1}{97.96}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)
HH
30 tháng 12 2017
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}=-\frac{9799}{9900}\)
29 tháng 6 2016
1) Tính:
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)\(-\frac{1}{2.1}\)
\(=\frac{1}{100}-\frac{1}{99}-\frac{1}{99}-\frac{1}{98}-\frac{1}{98}-\frac{1}{97}-\)\(...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)
\(=\frac{1}{100}-\frac{1}{1}\)
\(=-\frac{99}{100}\)
b) \(GọiB=\frac{-1}{100.99}+\frac{-1}{99.98}+...+\frac{-1}{2.1}\)
\(2B=\frac{-2}{100.99}+\frac{-2}{99.98}+...+\frac{-2}{2.1}\)
\(2B=\frac{-1}{100}-\frac{-1}{99}+\frac{-1}{99}-\frac{-1}{98}+...+\frac{-1}{2}-\frac{-1}{1}\)
\(2B=\frac{-1}{100}-\frac{-1}{1}\)
\(2B=\frac{99}{100}\Rightarrow B=\frac{99}{100}:2=\frac{99}{200}\)
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(< =>3A=\frac{3}{3}+\frac{3}{3^2}+\frac{3}{3^3}+...+\frac{3}{3^8}\)
\(< =>3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(< =>3A-A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
\(< =>A=\frac{3^8-1}{\frac{3^8}{2}}\)