Để tính tổng 11009×2016+11010×2015+…+12015×1010+11016×10091009×20161​+1010×20151​+…+2015×10101​+1016×10091​, ta có thể sử dụng một số kỹ thuật trong toán học. Trong trường hợp này, ta sẽ sử dụng tích phân.

Gọi $S$ là tổng cần tính, ta có thể viết nó dưới dạng tổng tỉ lệ:

�=11009×2016+11010×2015+…+12015×1010+11016×1009S=1009×20161​+1010×20151​+…+2015×10101​+

Sửa đề: Cho \(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)và \(Y=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\). Tính \(\frac{V}{Y}\)

\(V=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)

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

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

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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

\(=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

=> \(\frac{V}{Y}=\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}=1\)

V = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)

V = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)

V = \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

V = \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)

V = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)

V = \(\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\)

Vậy V : Y = \(\frac{\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}}{\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2016}}\)

( Mình nghĩ Y = 1/1009 + 1/1010 + ... + 1/2016 / Nếu Y như mình nói thì V : Y = 1 )

Sửa đề: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)

Ta có: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2013}-\frac{1}{2014}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac12+\frac14+\cdots+\frac{1}{2014}\right)\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-1-\frac12-\cdots-\frac{1}{1007}\)

\(=\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}\)

Ta có: \(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}\)

\(=\frac{2}{1008\cdot2014}+\frac{2}{1009\cdot2013}+\cdots+\frac{2}{1510\cdot1512}+\frac{1}{1511\cdot1511}\)

\(=2\left(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)

\(=\frac{2}{3022}\left(\frac{3022}{1008\cdot2014}+\frac{3022}{1009\cdot2013}+\cdots+\frac{3022}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)

\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\cdots+\frac{1}{1510}+\frac{1}{1512}\right)+\frac{1}{1511}\cdot\frac{1}{1511}\)

\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)\)

Ta có: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)

\(=\frac{\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)}\)

\(=1:\frac{1}{1511}=1511\)

Giải hộ mình đi

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1008}{1009}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1008}{1009}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1008}{1009}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1008}{1009}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1008}{1009}\)

\(\Leftrightarrow1-\frac{2}{x+1}=\frac{1008}{1009}\)

\(\Leftrightarrow\frac{-2}{x-1}=\frac{1008}{1009}-1\)

\(\Leftrightarrow\frac{-2}{x+1}=\frac{-1}{1009}\)

\(\Leftrightarrow-1.\left(x+1\right)=-2.1009\)

\(\Leftrightarrow-x-1=-2018\)

\(\Leftrightarrow-x=-2018+1=-2017\)

\(\Leftrightarrow x=2017\)

Vậy x=2017

Xét số chia: 1-\(\frac{1}{2}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) - \(\frac{1}{2016}\)

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

= (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{2015}\) + \(\frac{1}{2016}\)) - (1+\(\frac{1}{2}\) + \(\frac{1}{3}\) + \(\frac{1}{4}\) +...+\(\frac{1}{1007}\) + \(\frac{1}{1008}\))

=\(\frac{1}{1009}\) + \(\frac{1}{1010}\) + ... + \(\frac{1}{2015}\)+ \(\frac{1}{2016}\) => A=1

\(CMR:\dfrac{1}{2}+\dfrac{1}{3}.4+\dfrac{1}{5}.6+...+\dfrac{1}{2015}.2016=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2016}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{2015.2016}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2016}\left(đpcm\right)\).

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2014}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}-1-\frac{1}{2}-...-\frac{1}{1007}\)

\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2014}\)   (đpcm)