tính P=1/1.2+1/2.3+...+1/99.100
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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?
\(=\dfrac{1\cdot3}{1\cdot2}+\dfrac{1\cdot3}{2\cdot3}+...+\dfrac{1\cdot3}{99\cdot100}\\ =3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ =3\cdot\dfrac{99}{100}\\ =\dfrac{297}{100}\)
1/1.2 + 1/2.3 + ...... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + ...... + 1/99 - 1/100
= 1 - 1/100
= 99/100
1/1.2 + 1/2.3 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
b: Tổng của N là:
\(\dfrac{49\cdot48}{2}=49\cdot24=1176\)
Answer:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{100}{100}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+....+\frac{100}{99.100}-\frac{99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
1-1/2+1/2-1/3+1/3-1/4+1/4-..........+1/99-1/100=1-1/100=99/100
= 1 -1/2 + 1/2 - 1/3 +......+1/99 - 1/100
= 1 -1/100
= 99/100
***Ai k mk mk k lại !!***
1/1*2+1/2*3+1/3*4+...+1/99*100
=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/99-1/100)
=1-1/100
=99/100
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
ta có
1=1
1/1.2 =1-1/2
1/2.3 1/2-1/3
1/3.4 =1/3-1/4
.......
1/(99.100) =1/99 -1/100
cộng theo vế các đẳng thức trên được
S =1+1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
S =2-1/100
\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)