\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)

\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)

\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)

\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)

\(2A=2+3+4+5+6+...+2012+2013+2014\)

\(2A=\dfrac{\left(2+2014\right).2013}{2}\)

\(A=\dfrac{2016.2013}{4}=504.2013\)

\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)

\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)

\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)

\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)

\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)

\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)

\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)

\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\cdot x=\left(1+\dfrac{2011}{2}\right)+\left(1+\dfrac{2010}{3}\right)+...+\left(\dfrac{1}{2012}+1\right)+1\)

\(\Leftrightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2013}\)

=>x=2013

1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)

\(B< 1\)

2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)

\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)

\(B=\dfrac{1}{20}\)

3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)

\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)

\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)

\(A=11\)

4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)

Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)

\(\Rightarrow B>A\)

a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)

\(\Leftrightarrow\)\(2^x.15=480\)

\(\Rightarrow\)\(2^x=480:15\)

\(\Leftrightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5.

Số số hạng của B là 1914(là 1 số chẵn)

\(\Rightarrow B=\left(1-\dfrac{1}{2013^2}\right)\left(1-\dfrac{1}{2012^2}\right)\left(1-\dfrac{1}{2011^2}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{100^2}\right)\)

\(B=\dfrac{2013^2-1}{2013^2}\cdot\dfrac{2012^2-1}{2012^2}\cdot\dfrac{2011^2-1}{2011^2}\cdot\cdot\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)

\(B=\dfrac{2014\cdot2012\cdot2013\cdot2011\cdot2012\cdot2010\cdot...\cdot101\cdot99}{2013\cdot2013\cdot2012\cdot2012\cdot2011\cdot2011\cdot...\cdot100\cdot100}\)

\(B=\dfrac{2014\cdot99}{2013\cdot100}=\dfrac{3021}{3050}\)

Câu Q:Tương tự như mấy câu khác thôi

Bạn kiểm tra lại đề, \(f\left(x\right)=\dfrac{x^3}{1-3x-3x^2}\) hay \(f\left(x\right)=\dfrac{x^3}{1-3x+3x^2}\)

https://hoc24.vn/hoi-dap/question/598367.html

S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)

= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)

S=P nhé