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\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{1}-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
\(A=2017:\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(=2017:\left(1-\dfrac{1}{2018}\right)\)
\(=2017:\dfrac{2017}{2018}\)
\(=2017\cdot\dfrac{2018}{2017}\)
\(=2018\)
#NgDat
\(A=2017:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+...+\dfrac{1}{2017}\cdot\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{1}{1}-\dfrac{1}{2018}\right)\)
\(A=2017:\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)
\(A=2017:\dfrac{2017}{2018}\)
\(A=2018.\)
Ta có công thức :
\(\frac{1}{k\left(k+1\right)}=\frac{\left(k+1\right)-k}{k\left(k+1\right)}=\frac{k+1}{k\left(k+1\right)}-\frac{k}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}=\frac{n-1}{n}\)
Siêu tốc tổng quát: \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)áp vào
\(A=\frac{1}{1}-\frac{1}{14}=1-\frac{1}{14}\)
A=(2-1)/1.2+(3-2)/2.3+...+(14-13)/13.14
A=1-1/2+1/2-1/3+...+1/13-1/14
A=1-1/14=13/14
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
3A = 1.2.( 3 + 0 ) + 2.3.( 4 - 1 ) + .. + 99.100.( 101 - 98 )
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 99.100.101 - 98.99.100
3A = 99.100.101
A = ( 99.100.101 ) : 3 = 333300
Vậy A = 333300
mk làm câu b
A=1.2+2.3+3.4+.......+99.100
3.A =3.1.2+2.3.3+3.4.3+............+99.100.3
3.A= 1.2.3+2.3.(4-1)+3.4.(5-2) +..........+99.100.(101-98)
3.A=1.2.3+2.3.4-1.2.3 +3.4.5-2.3.4+............+99.100.101-98.99.100
vì cứ +2.3.4 lại -2.3.4 cứ như thế
3.A=99.100.101
A=(99.100.101):3
A=333300
chúc bạn may mắn trong học tập
mk vừa học xong
Trước tiên, chúng ta cần có lý thuyết về biến đổi phân số.
\(\dfrac{b-a}{a\cdot b}=\dfrac{1}{a}-\dfrac{1}{b}\)
Ta có:
\(S=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\)
\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(S=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+...-\dfrac{1}{2018}\)
\(S=1-\dfrac{1}{2018}\)
\(S=\dfrac{2017}{2018}\)
=1/1.2+1/2.3+1/3.4+...1/2017.2018
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018
=1-1/2018
=2018/2018-1/2018
=2017/2018