\(B=\dfrac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)

\(2B=\dfrac{2+2^2+2^3+...+2^{2009}}{1-2^{2009}}\)

\(B-2B=\)\(\dfrac{1+2+2^2+...+2^{2008}}{1-2^{2009}}\)\(-\dfrac{2+2^2+2^3+...+2^{2009}}{1-2^{2009}}\)

\(-B=\dfrac{1-2^{2009}}{1-2^{2009}}\)

B=-1

ta có:

2B = 2 + 2^2 +...+ 2^2009 / 1 - 2^2009

2B - B = (2 + 2^2 +...+ 2^2009)-(1 + 2 +...+ 2^2008) / 1 - 2^2009

B = 2^2009 - 1 / 1 - 2^2009

B = -(2^2009 - 1) / 1 - 2^2009    * (-1)

B = 1 * (-1)

B = -1

Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)

\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(\Rightarrow B=2.\frac{3}{16}\)

\(\Rightarrow B=\frac{3}{8}\)

Vậy \(B=\frac{3}{8}\)

C=220 +230 +242 +...+2240 =2×(120 +130 +142 +...+1240 )

C=2×(14×5 +15×6 +16×7 +...+115×16 )

 2/20 + 2/30 + 2/42 + 2/56 + .... + 2/240

                   = 2/4x5 + 2/5x6 + 2/6x7 + 2/7x8 + ... + 2/15x16

                   = 2 x (1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/15 - 1/16)

                   = 2 x (1/4 - 1/16)

                   = 2 x 3/16 = 3/8 

k nha

\(B=1+\dfrac{1}{2}\cdot\dfrac{\left(1+2\right)\cdot2}{2}+\dfrac{1}{3}\cdot\dfrac{\left(1+3\right)\cdot3}{2}+...+\dfrac{1}{20}\cdot\dfrac{\left(20+1\right)\cdot20}{2}\\ B=1+\dfrac{3}{2}+2+\dfrac{5}{2}+...+10+\dfrac{21}{2}\\ B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{20}{2}+\dfrac{21}{2}\\ B=\dfrac{2+3+...+20+21}{2}=\dfrac{\dfrac{\left(21+2\right)\cdot20}{2}}{2}=\dfrac{23\cdot10}{2}=115\)

\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)

\(B=\frac{\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)

\(B=\frac{\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}}\)

\(B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}\)

\(B=2016\)

\(B=\frac{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+...+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)

\(\Rightarrow B=\frac{1+\left(\frac{2014}{2}+1\right)+\left(\frac{2013}{3}+1\right)+\left(\frac{2012}{4}+1\right)+...+\left(\frac{1}{2015}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)

\(\Rightarrow B=\frac{\frac{2016}{2016}+\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)

\(\Rightarrow B=\frac{2016\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}\)

\(\Rightarrow B=2016\)

Vậy \(B=2016\)

Có \(B=\frac{\frac{2015}{1}+\frac{2014}{2}+......+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2016}}\)

Xét mẫu số:

Đặt A là mẫu số; C là tử số

\(A=\frac{2015}{1}+\frac{2014}{2}+......+\frac{1}{2015}\)

\(=\left(\frac{1}{2015}+1\right)+\left(\frac{2}{2014}+1\right)+.........+\left(\frac{2015}{1}-2014\right)\)

\(=\frac{2016}{2015}+\frac{2016}{2014}+.........+\frac{2016}{2016}\)

\(=2016.\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+..........+\frac{1}{2}\right)\)

\(=2016.C\)

\(\Rightarrow B=\frac{2016.C}{C}=2016\)

Vậy B = 2016

Tham khảo:M=1/3+1/3^2+1/3^3+....+1/3^99. CMR: M<1/2

Ta có:1/(3^n)+1/(3^(n+1))=2/(3^(n+1))(cái này bạn tự quy đồng ra ra nhé!). 
Áp dụng ta có:1-1/3=2/3 
1/3-1/(3^2)=2/(3^2) 
1/(3^2)-1/(3^3)=2/(3^3) 
.... 
1/(3^98)-1/(3^99)=2/(3^99). 
Cộng từng vế các phép tính với nhau ta có:1-1/(3^99)=2M. 
Mà 1-1/(3^99)<1 nên 2M<1 nên M<1/2(điều phải chứng minh)

Bạn Thiên bình có 102 sai đề bài rồi

Với cả chép trên mạng nữa

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

\(3B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

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

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

\(4B=1-\frac{1}{3^{100}}\)

=> \(B=\frac{1-\frac{1}{3^{100}}}{4}\)

\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

=3/4*4/5*...*99/100

=3/100