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a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
Mình gõ câu a bị lỗi nha , thực chất câu a là
a) Tìm các số tự nhiên x, y biết : 2xy + x + 2y = 13
a)Bạn làm nha vì bài này dễ rồi
b)+)Ta có:A=1.2+2.3+3.4+..................+99.100
=>3A=1.2.3+2.3.3+3.4.3+.................+99.100.3
=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)
=>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101
=>3A=99.100.101
=>A=\(\frac{99.100.101}{3}=333300\)
+)Ta lại có:B=12+22+32+..................+992
=>B=1.1+2.2+3.3+............+99.99
=>B=1.(2-1)+2.(3-1)+3.(4-1)+..........+99.(100-1)
=>B=1.2-1+2.3-2+3.4-3+........................+99.100-99
=>B=(1.2+2.3+3.4+............+99.100)-(1+2+3+..............+99)
Đặt N=1.2+2.3+3.4+....................+99.100
=>3N=1.2.3+2.3.3+3.4.3+.................+99.100.3
=>3N=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)
=>3N=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101
=>3N=99.100.101
=>N=\(\frac{99.100.101}{3}=333300\)
Đặt M=1+2+3+..............+99(có 99 số hạng)
=>M=\(\frac{\left(1+99\right).99}{2}=4950\)
+)Ta thấy A-B=333300-(333300-4950)
=>A-B=333300-333300+4950
=>A-B=4950\(⋮\)50
Vậy A-B\(⋮\)50
Chúc bn học tốt
Ta có:
\(A=\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{1000}\right)^2< 1\)
\(A=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{1000000}< 1\)
\(\frac{1}{4}< \frac{1}{1\cdot2}\)
\(\frac{1}{9}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1000000}< \frac{1}{999.1000}\)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{999\cdot1000}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(A< \frac{1}{1}-\frac{1}{1000}< 1\)
\(\Rightarrow A< 1\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{999.1000}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{999}-\frac{1}{1000}\)
\(A< 1-\frac{1}{1000}\)
\(=>A< 1\)
\(=>ĐPCM\)
(1.2 + 2.3 + 3.4 + ... + 2018.2019) - (12 + 22 + ... + 20182)
= (1.2 + 2.3 + ... + 2018.2019) - (1.1 + 2.2 + ... + 2018.2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.(2 - 1) + 2.(3 - 1) + ... + 2018.(2019 - 1)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019 - 1 - 2 - 3 - ... - 2018)
= (1.2 + 2.3 + ... + 2018.2019) - [1.2 + 2.3 + ... + 2018.2019 - (1 + 2 + ... + 2018)]
= (1.2 + 2.3 + ... + 2018.2019) - (1.2 + 2.3 + ... + 2018.2019) + (1 + 2 + 3 + ... + 2018)
= 1 + 2 + ... + 2018 (có : (2018 - 1) : 1 + 1 = 2018 (số))
= (2018 + 1).2018 : 2
= 2037171
A = 1+2+22+...+210
=> 2A = 2+22+23+...+211
=> 2A - A = (2+22+23+...+211) - (1+2+22+...+210)
=> A = 211 - 1
B = 1+3+32+...+310
=> 3B = 3+32+33+...+311
=> 3B - B = (3+32+33+...+311) - (1+3+32+...+310)
=> 2B = 311 - 1
=> B = \(\frac{3^{11}-1}{2}\)
A = 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10
2A = 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11
2A - A = ( 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11 )
- ( 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10 )
A = 2 11 - 1
A = 2047
B = 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10
3B = 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11
3B - B= ( 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11 )
- ( 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10 )
2B = 3 11 - 1
B = \(\frac{3^{11}-1}{2}\)
B = 88573
a: 6S=6+6^2+...+6^65
=>5S=6^65-1
=>S=(6^65-1)/5
b: 4S=4+4^2+...+4^401
=>3S=4^101-1
=>S=(4^101-1)/3
c: 9S=3^2+3^4+...+3^104
=>8S=3^104-1
=>S=(3^104-1)/8