Tính : \(B=\dfrac{1}{1.100}+\dfrac{1}{2.99}+\dfrac{1}{3.98}+\dfrac{1}{4.97}+...+\dfrac{1}{50.51}\)
K
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KQ
2
11 tháng 6 2018
\(F=1.100+2.99+3.98+...+99.2+1.100\)
\(=1.100+2.\left(100-1\right)+3.\left(100-2\right)+....+100.\left(100-99\right)\)
\(=1.100+2.100-2.1+3.100-3.2+...+100.100-100.99\)
\(=100.\left(1+2+3+...+100\right)-\left(1.2+2.3+3.4+....99.100\right)\)
\(=100.\frac{101.100}{2}-\frac{99.100.101}{3}=505000-333300=171700\)
Đúng thì ủng hộ nha!
TN
11 tháng 6 2018
F = 1.100 + 2.(100-1) + 3. (100-2) +...+100.(100-99)
F = 1.100 + 2.100 - 1.2 + 3.100 - 2.3 + ...+ 100.100 - 99.100
F = 100 . ( 1+2+3+...+100) - (1.2+2.3+3.4+...+99.100)
= 100.(101.100:2) - (99.100.101:3) = 505000 - 333300 = 17100
PH
1
16 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
a.
$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$
$=1-\frac{1}{1000}=\frac{999}{1000}$
AH
Akai Haruma
Giáo viên
15 tháng 4 2023
b.
$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$
$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$
$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$
$=1-\frac{1}{500}=\frac{499}{500}$
$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$
NA
Nguyễn An Ninh
VIP
22 tháng 4 2023
A=20/1.21+20/2.22+...+20/80.100
=1-1/21+1/2-1/22+...+1/80-1/100
=(1+1/2+...+1/80)-(1/21+1/22+...+1/100)
80B=80/1.81+80/2.82+...+8/20.100
=1-1/81+1/2-1/82+...+1/20-1/100
=(1+1/2+...+1/20)-(1/81+1/82+...+1/100)
=(1+1/2+1/3+...+1/20+1/21+1/22+...+1/80)-(1/21+1/22+...1/80+1/81+1/82+...1/100)
=>20A=80B
=>A=4B
AH
Akai Haruma
Giáo viên
20 tháng 9 2021
Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)
\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)
\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)
Mặt khác:
$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$
$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$
$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$
$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$
Từ $(1);(2)\Rightarrow 299A=101B$
$\Rightarrow \frac{A}{B}=\frac{101}{299}$
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 2 2021
Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
Trong tử số, có số số 1 là:
(100 - 1) + 1 = 100(số)
Trong tử số, có số số 2 là:
(100 - 2) + 1 = 99(số)
Trong tử số, có số số 3 là:
(100 - 3) + 1 = 98(số)
.........................................................................
Trong tử số, có số số 100 là:
(100 - 100) + 1 = 1(số)
Vậy, ta có:
1+(1+2)+(1+2+3)+...+(1+2+3+4+...+100)100.1+99.2+98.3+...+3.98+2.99+1.1001+(1+2)+(1+2+3)+...+(1+2+3+4+...+100)100.1+99.2+98.3+...+3.98+2.99+1.100
=100.1+99.2+98.3+...+3.98+2.99+1.100100.1+99.2+98.3+...+3.98+2.99+1.100=100.1+99.2+98.3+...+3.98+2.99+1.100100.1+99.2+98.3+...+3.98+2.99+1.100
=1