A=4/3+9/8+16/15+..............+4064256/4064255

A=1+1/3+1+1/8+1/15+...............+1/4064255

A=(1+1+...+1)+(1/3+1/8+...+1/406255)          (có 2015 số 1)

A=2015+(1/1.3+1/2.4+...........+1/2015.2017)
A=2015+1/2(1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7+....+1/2012-1/2014+1/2013-1/2015+1/2014-1/2016+1/2015-1/2017)

A=2015+1/2(1+1/2-1/2016-1/2017)

A=2015,749504

                                k cho mình nhé mình k lại cho

Trong dấu ngoặc đơn có số các số hạng là

Đặt tổng các số hạng trong ngoặc đơn là A

\(\dfrac{2013-1}{2}+1=1007\) số hạng

\(A=\dfrac{3+1}{1.3}-\dfrac{5+3}{3.5}+\dfrac{7+5}{5.7}-...+\dfrac{2015+2013}{2013.2015}=\)

\(=1+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{2013}+\dfrac{1}{2015}=1+\dfrac{1}{2015}=\dfrac{2016}{2015}\)

\(\Rightarrow M=A.\dfrac{2015}{2016}=\dfrac{2016}{2015}.\dfrac{2015}{2016}=1\) là số tự nhiên

\(\sum\limits^{2016}_{x=1}\left(\dfrac{x^2}{\left(2x-1\right)\left(2x+1\right)}\right)\)

Đặt A = \(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2010.2012+2}{2011^2}+\frac{2015.2017+2}{2016^2}\)

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

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

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

\(=\left(1+1+...+1\right)+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2016^2}\right)\)(2015 hạng tử 1)

\(=2015+\left(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{2016.2016}\right)\)

 \(< 2015+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)

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

= 2015 + 1 + 1/2016

= 2016 + 1/2016 < 2017 

=> A < 2017 (ĐPCM)

<=>2-2/3+2/3-2/5........+2n-2n+2<2015/2016

<=>2-2n+2<2015/2016

=>n+2=1/2016

=>n=2014

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{n\left(n+2\right)}\)<\(\frac{2015}{2016}\)

VT=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{5}-\frac{1}{n+2}\)=\(1-\frac{1}{n+2}\)

Ta có:\(1-\frac{1}{n+2}=\frac{2015}{2016}\Rightarrow\)\(\frac{1}{n+2}=1-\frac{2015}{2016}\)

\(\Rightarrow\)\(\frac{1}{n+2}=\frac{1}{2016}=n+2=2016\)

\(\Rightarrow\)\(n=2014\)

Vậy\(n=2014\)

Bài này dễ ý mà, vô cùng đơn giản..........

Ta có:

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}.\)

\(\dfrac{2}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)

\(1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}.\)

\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)

\(0+0+...+0+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)

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

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

\(\dfrac{1}{x+2}=\dfrac{1}{2016}.\)

\(\Rightarrow x+2=2016.\)

\(\Rightarrow x=2016-2=2014.\)

Vậy \(x=2014.\)

~ Học tốt nha bn!!! ~

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