Đố 1+1.2 =
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=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3
=> 3A = 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + .... + 49.50.( 51 - 48 )
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 49.50.51 - 48.49.50
=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + ( 48.49.50 - 48.49.50 ) + 49.50.51
=> 3A = 49.50.51
=> A = ( 49.50.51 ) : 3
=> A = 41650
A = 1.2 + 2.3 + 3.4 + ..... + 49.50
3A=1.2.3+2.3.3+3.4.3+...+49.50.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+48.49.(50-47)+49.50.(51-48)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+48.49.50-47.48.49+49.50.51-48.49.50
3A=(1.2.3-1.2.3)+(2.3.4-2.3.4)+...(47.48.49-47.48.49)-(48.49.50-48.49.50)+49.50.51
3A=0+0+...+0+0+49.50.51
3A=49.50.51
A=\(\frac{49.50.51}{3}\)
A=41650
Đáp số: A=41650

S=1.2+ 2.3+.......+99.100
Nhân cả 2 vế với 3, ta được:
3S=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
----> S = (99.100.101):3
S = 333300
Vậy S=333300

S=1.2+2.3+3.4+4.5+....+99.100
3S=1.2.3+2.3.3+3.4.3+4.5.3+....+99.100.3
3S=1.2.3+2.3.(4-1)+3.4.(5-1)+4.5(6-3)+....+99.100(101-98)
3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-2.4.5+....+99.100.101-98.99.100
3S=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+.......+99.100.101
3S=99.100.101
3S=999900
S=999900:3
S=333300
S=1.2+2.3+3.4+4.5+...+99.100
3S=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
3S=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
3S=99.100.101
S=(99.100.101):3=333300

a)
`1/1-1/2`
`=2/2-1/2`
`=1/2`
b)
`1/(1*2)+1/(2*3)`
`=1/1-1/2+1/2-1/3`
`=1/1-1/3`
`=3/3-1/3`
`=2/3`
c)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{1}-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
d)
\(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+...+\dfrac{3}{99\cdot100}\) đề phải như thế này chứ nhỉ?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)


\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(B=1.2+2.3+3.4+...+49.50\)
\(3B=1.2.3+2.3.3+3.4.3+...+49.50.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)
\(=49.50.51\)
\(B=\frac{49.50.51}{3}=49.50.17\)
\(50^2.A-\frac{B}{17}=49.50-49.50=0\)
bằng 3
Trả lời :
1+1.2=3
Hok_Tốt
#Thiên_Hy
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