1/1x2+1/2x3+1/3x4+...+1/2013x2014+1/2014x2015
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vì 1/1*2=1-1/2
1/2*3=1/2-1/3
.....................
1/2014*2015=1/2014-1/2015
=1-1/2+1/2-1/3+1/3-....+1/2014-1/2015
=1-1/2015
=2014/2115
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{2014x2015}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
A=1*2+2*3+...+2014*2015
3A=1*2*3+2*3*(4-1)+...+2014*2015*(2016-2013)
3A=1*2*3+2*3*4-1*2*3+...+2014*2015*2016-2013*2014*2015
3A=2014*2015*2016
A=2014*2015*2016/3
A=2727117120
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 2014x2015
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 2014x2015x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 2014x2015x(2016-2013)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 2014x2015x2016 - 2014x2015x2013.
A x 3 = 2014x2015x2016
A = 2014x2015x2016 : 3
A = 2727117120
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
Ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
\(=\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=\)\(1-\frac{1}{2014}\)
\(=\)\(\frac{2014}{2014}-\frac{1}{2014}\)
\(=\)\(\frac{2013}{2014}\)
Vậy \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}=\frac{2013}{2014}\)
Dấu \(.\) là dấu nhân nhé
Chúc bạn học tốt ~
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2013\times2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)
CHÚC BN HỌC TỐT!!!!!
1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2013*2014 + 1/2014*2015
= 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2013 - 1/2014 + 1/2014 - 1/2015
=1-1/2015
=2014/2015
=1-1/2+1/2-1/3+...+1/2014-1/2015
=1-1/2015
=2014/2015