cho A=\(\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+4031}\).so sanh A voi \(\frac{2015}{2016}\)
Những câu hỏi liên quan
chứng minh \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4031}{2015^2.2016^2}< 1\)
Cho A=1+ \(\frac{1}{3}\)+\(\frac{1}{5}\)+........+\(\frac{1}{2015}\) B =\(\frac{1}{2}\)+\(\frac{1}{4}\)+.........+\(\frac{1}{2016}\)
So sanh A voi B+1
CHỨNG MINH RẰNG : \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4031}{2015^2.2016^2}< 1\)
Cho A= \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+..............+\frac{11}{5^{12}}\)so sanh A voi \(\frac{1}{16}\)
cho A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2016}+\frac{1}{2017}\)
va B=\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+......+\frac{2}{2015}+\frac{1}{2016}\)
Tinh ti so \(\frac{A}{B}\)
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right);B=-\frac{1}{2}\). hãy so sanh a va b
\(A=\left(\frac{1}{1^2}-1\right)\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)\)
\(=0.\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2015^2}-1\right)\left(\frac{1}{2016^2}-1\right)=0>-\frac{1}{2}\)
suy ra A>B
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Cho :
A = \(\left(\frac{1}{2}_{ }+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2016}+\frac{1}{2017}\right)\)
B = \(\left(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Tính \(\frac{B}{A}\) ?
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sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
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\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
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A =\(\frac{2015+\frac{2014}{2}+\frac{2013}{3}+\frac{2012}{4}+\frac{2011}{5}+.....+\frac{1}{2015}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2016}}=\)
tìm A
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
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cho A=1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026},B=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\)so sanh \(\frac{A}{B}\)voi\(1\frac{2013}{2014}\)