\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{14.15}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{14}-\frac{1}{15}\)

\(A=1-\frac{1}{15}\)

\(A=\frac{14}{ }15\)

=1-1/2+1/2-1/3+...+1/14-1/15

=1-1/15

=14/15

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{14.15}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{14}-\frac{1}{15}\right).\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}=1-\left(\frac{1}{2}-\frac{1}{2}\right)_{ }-...-\left(\frac{1}{14}-\frac{1}{14}\right)-\frac{1}{15}_{ }.\)

\(=1-\frac{1}{15}=\frac{14}{15}.\)

=1x2x3+2x3x3+...+98x99x3

=1x2x3+2x3x(4-1)+...+98x99x(100-97)

=1x2x3+2x3x4-1x2x3+...+98x99x100-97x98x99

=98x99x100

=970200

1/1x2+1/2x3+1/3x4+..+1/9x10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...-1/10

=1-1/10

=9/10

\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)

\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)

\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)

\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)

\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)

\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)

3C=1x2x3+2x3x3+3x4x3+...+Xx(X+1)=

=1x2x3+2x3x(4-1)+3x4x(5-2)+...+Xx(X+1)[(X+2)-(X-1)]=

=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-(X-1)xXx(X+1)+Xx(X+1)x(X+2)=

=Xx(X+1)(X+2)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}\)

\(=\frac{2010}{2011}\)

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011

= 1 - 1/2011

= 2010/ 2011

Đáp số: 2010/2011

Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2010 - 1/2011

= 1 - 1/2011

= 2010/2011

Đáp sô: 2010/2011

Chú ý công thưc:              \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(B=\dfrac{3}{1x2}+\dfrac{3}{2x3}+...+\dfrac{3}{50x51}\)

\(B=3x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{50x51}\right)\)

\(B=3x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)

\(B=3x\left(1-\dfrac{1}{51}\right)=3x\dfrac{50}{51}=\dfrac{150}{51}\)

CẢM ƠN BN

\(\frac{1}{1x2}+\frac{1}{1x3}+...+\frac{1}{999x1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

1/1x2+1/2x3+1/3x4+...+1/999x1000

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000

=1-1/1000

=1000/1000-1/1000

=999/1000

999 / 1000

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

Đặt A = 1x2+2x3+3x4+...+25x26+26x27

3A = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4  -2 x 3 x 4 + ..... + 26 x 27 x 28

3A = 26 x 27 x 28

A= \(\text{ }\frac{\text{26 x 27 x 28}}{3}=6552\)

Cậu học cấp 2 rồi còn giải bài cấp 1 làm chi.

giải chi tiết giúp mình