Tính
C= (1-1/2)+(1-1/6)+....+(1-1/9900)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KK
1
XO
21 tháng 7 2019
C = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{9900}\right)\)(99 CẶP)
= \(\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)(99 SỐ HẠNG 1)
= \(1.99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
= \(99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(99-\left(1-\frac{1}{100}\right)\)
= \(99-\frac{99}{100}\)
= \(\frac{9801}{100}\)
Vậy \(C=\frac{9801}{100}\)
Chúc bạn học tốt !!!!!
DT
2
TH
Thầy Hùng Olm
Manager
VIP
13 tháng 4 2023
A = 1+ 1+1+ ...+ 1 +(\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}+\dfrac{1}{10100}\))
=(1+1+1+...+1)+ (\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}+\dfrac{1}{100x101}\))
=100 +\(1-\dfrac{1}{101}=100-\dfrac{100}{101}=\dfrac{10000}{101}\)
LH
13 tháng 4 2023
1+1/2+1+1/6+1+1/12+...+1+1/9900
=1+1/1*2+1+1/2.3+....+1+1/99*100
=100*1+1-1/2+1/2-1/3+1/3-1/4...+1/99-1/100
=100+99/100
=10099/100
LT
3
25 tháng 6 2016
\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)...\left(1-\frac{2}{9900}\right)\)
\(=\frac{4}{6}.\frac{10}{12}...\frac{9898}{9900}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{98.101}{99.100}\)
\(=\frac{1.2...98}{3.4...100}.\frac{4.5...101}{2.3...99}\)
\(=\frac{2}{99.100}.\frac{100.101}{2.3}\)
\(=\frac{101}{99.3}\)
\(=\frac{101}{297}\)
MT
1
NQ
1
LQ
4 tháng 1 2015
T= 1 - 1/2 + 1/2 - 1/3 + ......+ 1/99 - 1/100
= 1 - 1/100
= 99/100
LT
3
23 tháng 10 2015
\(t=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(t=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(t=1-\frac{1}{100}=\frac{99}{100}\)
Vậy \(t=\frac{99}{100}\)
QM
2
TM
3 tháng 5 2023
a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)
b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).
Vậy : \(S=\dfrac{99}{100}.\)
NB
3 tháng 5 2023
a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)
\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)
b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)