CMR:
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2015^2+2016^2}
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PC
12 tháng 4 2016
TA CÓ \(\left(X-Y\right)^2\ge0\Rightarrow X^2-2\cdot X\cdot Y+Y^2\ge0\Rightarrow X^2+Y^2\ge2\cdot X\cdot Y\) \(\Rightarrow\frac{1}{5}=\frac{1}{1^2+2^2}<\frac{1}{2}\cdot\frac{1}{1\cdot2}\)
TƯƠNG TỰ TA CÓ \(\frac{1}{13}<\frac{1}{2}\cdot\frac{1}{2\cdot3}\) ................\(\frac{1}{2015^2+2016^2}<\frac{1}{2}\cdot\frac{1}{2015\cdot2016}\)
\(\Rightarrow\) \(\frac{1}{5}+\frac{1}{13}+..........+\frac{1}{2015^22016^2}<\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.........+\frac{1}{2015\cdot2016}\right)=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2016}\right)\)
VÌ \(1-\frac{1}{2016}<1\Rightarrow\frac{1}{2}\cdot\left(1-\frac{1}{2016}\right)<\frac{1}{2}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+....+\frac{1}{2015^2+2016^2}<\frac{1}{2}\)
16 tháng 11 2017
Ta có :
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)là số nguyên
TD
3
NV
0
Nhận xét: Với 2 số a; b bất kì ta có (a - b)2 \(\ge\) 0 => a2 - 2ab + b2 \(\ge\) 0 => a2 + b2 \(\ge\) 2ab
Áp dụng ta có: 5 = 12 + 22 \(\ge\) 2.1.2
13 = 22 + 32 \(\ge\) 2.2.3
25 = 32 + 42 \(\ge\) 2.3.4
..........
20152 + 20162 \(\ge\) 2.2015. 2016
=> \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2015^2+2016^2}\le\frac{1}{2.1.2}+\frac{1}{2.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2.2015.2016}\)
=> \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2015^2+2016^2}\le\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
=> \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+...+\frac{1}{2015^2+2016^2}\le\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\le\frac{1}{2}\left(1-\frac{1}{2016}\right)