CMR: A=1/2+1/2^2+1/2^3+...+1/2^2016+1/2^2017<1
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Những câu hỏi liên quan
cmr 1/(1^2+2^2)+1/(2^2+3^2)+...+1/(2016^2+2017^2) <1/2
1/2+3/4+5/6+...+2015/2016 CMR A^2<1/2017
CMR: \(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)là 1 số hữu tỉ
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
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Cho P= 12017+22017+32017+...+20162017 , Q= 1+2+3+...2016
CMR: P chia hết cho Q
A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
CMR:
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)là 1 số hữu tỉ
bạn chứng minh bài toán tổng quát : \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)rồi áp dụng vào giải bài này nhé
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Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017};B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\).CMR B/A là số nguyên
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
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CMR
\(\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{2016^2+2017^2}< \dfrac{1}{2}\)
(a - b)2 \(\ge0\Leftrightarrow a^2+b^2-2ab\ge0\Leftrightarrow a^2+b^2\ge2ab\)
=> \(\frac{1}{a^2+b^2}< \frac{1}{2ab}\left(a;b>0;a\ne b\right)\)
Áp dụng vào bài toán ta có:
\(\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+\frac{1}{3^2+4^2}+...+\frac{1}{2016^2+2017^2}< \frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(< \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(< \frac{1}{2}\left(1-\frac{1}{2017}\right)< \frac{1}{2}\left(đpcm\right)\)
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K-2016=1+ (1+2)+(1+2+3)+….+(1+2+3+…+2017)/2017*1+2016*2+2015*3+…+2*2016+1*2017
tìm K