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S = 1.2 + 2.3 + 3.4 + ... + 99.100
3S = 1.2.3 + 2.3.3 + 3.4.3 + .. +99.100.3
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3S = 99.100.101
3S = 999 900
S = 333 300
Vậy S = 333 300
A= 1-2+3-4+4-5+...+99-100
A = ( 1 - 2 ) + ( 2 - 3 ) + ....+ ( 99 - 100 )
A = ( - 1 ) + ( - 1 ) +....+ ( - 1 )
A = ( - 1 ) . 50
A = - 50
B = 1.2 + 2.3 + 3.4 + 4.5 +...+ 99.100
Nhân cả 2 vế với 3, ta được:
3A=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
=) B = (99.100.101) :3
B = 333300
Vậy B= 333300
A= 1-2+3-4+4-5+...+99-100
A = (1-2) + (3-4) + (4-5) + ... + (99-100)
A = (-1) + (-1) + (-1) + ...+ (-1)
A = (-1).50
A = 1
\(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2017.\left(1-\frac{1}{100}\right)\)
\(=2017.\frac{99}{100}\)
\(=\frac{199693}{100}\)
\(\frac{2017}{1.2}+\frac{2017}{3.4}+\frac{2017}{4.5}+...+\frac{2017}{99.100}\) \(=2017.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\) \(=2017.\left(1-\frac{1}{100}\right)\) \(=2017.\frac{99}{100}\) \(=\frac{199693}{100}\)
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)
Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath
= 1/1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 +.....+1/99 + 1/100
=( 1/1 + 1/2 +1/3 +1/4 + 1/5 + 1/6 +.....1/99 + 1/100) - 2(1/2 + 1/4 + 1/6 + .....+ 1/100)
=(1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +.....+ 1/99 + 1/100) - ( 1 + 1/2 + 1/3 + .... + 1/50)
= 1/51 + 1/52 + 1/53 +....+ 1/100....>1/100
= ( 1/51 + 1/52 + 1/53 +.....+ 1/75) + ( 1/76 + 1/77 + 1/78 +.....+ 1/100)
Có 1/51>1/52>1/53>....>1/75 ; 1/76>1/77>1/78>....>1/100
A> 1/75.25 + 1/100.25= 1/3 + 1/4 = 7/12
A< 1/51.25+ 1/76.25 < 1/50.25 + 1/75.25= 1/2+1/3=5/6
Vậy 7/12< A< 5/6
A = 1.2 + 2.3 + 3.4 + .. + 99.100
<=> 3A = 1.2.3 + 2.3.3 + 3.4.3 +...+ 99.100.3
= 1.2.3 + 2.3.(4-1) + 3.4.( 5 -2) +...+ 99.100.(101-98)
= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 + ..- 98.99.100 + 99.100.101
= 999900
<=> A = 999900 : 3 = 333300
A=1.2+2.3+3.4+...+99.100
3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3A=1.2.3+2.3.4+3.4.5+...+98.99.100+99.100.101 - 0.1.2-1.2.3-2.3.4-3.4.5-...-98.99.100
3A=99.100.101-0.1.2
3A=999900-0
3A=999900
A=999900:3
A=333300
A=1.2+ 2.3+.......+99.100
Nhân cả 2 vế với 3, ta được:
3A=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
----> A = (99.100.101):3
A = 333300
Vậy A=333300