giúp vs

b) 

A=1.2+2.3+3.4+...+2010.2011

3A=1.2.3+2.3.3+3.4.3+...+2010.2011.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2010.2011.(2012-2009)

=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-2009.2010.2011+2010.2011.2012

=2010.2011.2012

=>A=2010.2011.2012 / 3

=2710908440

Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010}\)

\(\Rightarrow x+1=2010\)

\(\Rightarrow x=2010-1\)

\(\Rightarrow x=2009\)

Vậy x = 2009

=> 1-1/2+1/2-1/3+1/3- 1/4 +... +1/x -1/x+1 = 2009/1020

=> 1 - 1/x+1=2009/2010

=> (x+1-1)/x+1=2009/2010

=> x/x+1=2009/2010

=>x=2009

b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1

b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)

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

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

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

c,Ta thấy

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

\(\frac{1}{4^2}< \frac{1}{3.4}\)

\(.....\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

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

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)

Lời giải :

Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101

3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3

=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)

=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102

=100.101.102

S=100.101.34=343400

1.Tính 

a) Ta có: 

  A=(1-1/2²).(1-1/3²)...(1-1/100²)

=>A=3/2².8/3².....9999/100²

=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)

=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)

=>A=1/100.101/2

=>A=101/200

b) Ta có: 

  B=-1/1.2-1/2.3-1/3.4-...-1/100.101

=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)

=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)

=>B=-(1-1/101)

=>B=-100/101

 c) Ta có:

 C=1.2+2.3+3.4+...+100.101

       =>3C=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3C=100.101.102

       =>3C=1030200

       =>C=343400

Chúc bạn hok tốt nhé >:)!!!!!

trong sách nâng cao và phát triển có đó bạn

1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 +…+ 2009 – 2010 – 2011 + 2012.
= (1 – 2 – 3 + 4) +( 5 – 6 – 7 + 8) +…+ (2009 – 2010 – 2011 + 2012)
= 0 + 0 + ...+ 0 = 0

bai toan qua kho

\(S=-\left(1+2+...+2^{2009}+2^{2010}\right)\)

\(-2S=2\left(1+2+...+2^{2009}+2^{2010}\right)\)

\(\Rightarrow-2S+S=-S=2+2^2+...+2^{2010}+2^{2011}-1-2-...-2^{2009}-2^{2010}\)

\(-S=2^{2011}-1\Rightarrow S=1-2^{2011}\)

S=2²⁰¹⁰ - 2²⁰⁰⁹ - 2^{2008 -...-2-1}

=>2S=2 x 2²⁰¹⁰ - 2 x 2²⁰⁰⁹ - 2 x 2²⁰⁰⁸ -...-2 x 2 -2 x 1

2S=2²⁰¹¹ - 2²⁰¹⁰ - 2²⁰⁰⁹ - ... - 2² -2

=>S=1-2²⁰¹¹