A=1(2+1)+2(3+1)+3(4+1)+...+99(100 +1 )

A=1.2+1+2.3+2+3.4+3...99.100+99

A=(1.2+2.3+3.4+...99.100)+(1+2+3+4...99)

giải:

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

24497550

\(2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2S=\frac{1}{2}-\frac{1}{9900}\)

\(2S=\frac{4949}{9900}\)

\(S=\frac{4949}{19800}\)

Ta xét : \(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4}\)

...

\(\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)

Ta có : 2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

=> 2S = \(\frac{1}{1.2}-\frac{1}{99.100}\)

=> 2S = \(\frac{4949}{9900}\)

=> S = \(\frac{4949}{19800}\)

2S=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\)

2S= \(1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)2S= 1- \(\dfrac{1}{100}\)

2S= \(\dfrac{99}{100}\)

S= \(\dfrac{99}{100}.\dfrac{1}{2}\)

S=\(\dfrac{198}{100}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+....+\frac{1}{98.99.100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(=\frac{1}{2}.\frac{4949}{9900}\)

\(=\frac{1}{19800}\)

Nhầm , kết quả bằng :

\(=\frac{4949}{19800}\)

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

TICK ĐÚNG GIÚP MÌNH Ặ

Bài 2 :

\(B=2014\cdot2020\)

\(B=\left(2017-3\right)\left(2017+3\right)\)

\(B=2017^2-3^2\)

\(B=2017^2-9< A=2017^2\)

Vậy \(B< A\)

\(B=2014.2020\)

\(B=\left(2017-3\right)\left(2017+3\right)\)

\(B=\left(2017-3\right).2017+\left(2017+3\right).3\)

\(B=2017^2-3.2017+2017.3+3^2\)

\(B=2017^2-3^2< 2017^2=A\)

  Vậy A > B

   _Hok tốt_

!!!

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+....+\frac{1}{98.99.100}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{99}{100}\)

=24497550

2A=\(\frac{2}{1.2.3}\)+\(\frac{2}{2.3.4}\)+\(\frac{2}{4.5.6}\)+...+\(\frac{2}{98.99.100}\)

2A=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\)+..+\(\frac{1}{98.99}\)-\(\frac{1}{99.100}\)

2A=\(\frac{1}{1.2}\)-\(\frac{1}{99.100}\)=\(\frac{1}{2}\)-\(\frac{1}{9900}\)=\(\frac{4949}{9900}\)

A=\(\frac{4949}{9900}\):2

A=\(\frac{4949}{19800}\)

= 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + 1/4.5 - ........+1/98.99 - 1/99.100 )

=1/2.(1/1.2 - 1/99.100)

=1/2 . 4949/9900

=4949/19800