Tính tổng \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2017}{2017^4+2017^2+1}\)
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Tính tổng:
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016+2017}\)
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)
\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)
\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)
\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)
Vậy \(S=\frac{1008}{1009}\)
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\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016+2017}\)
\(=\frac{1}{2\left(2+1\right):2}+\frac{1}{3\left(3+1\right):2}+\frac{1}{4\left(4+1\right):2}+....+\frac{1}{2017\left(2017+1\right):2}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2017.2018}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2018}\right)=2\cdot\frac{504}{1009}=\frac{1008}{1009}\)
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Tính : \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+...+\frac{2017}{2017^4+2017^2+2017}\)
Tính S = 1 + \(\frac{1}{2}\).(1+2) + \(\frac{1}{3}\).(1+2+3) + \(\frac{1}{4}\).(1+2+3+4) + .... + \(\frac{1}{2017}\).(1+2+3+....+2017)
tính
A=\(\frac{\frac{2017}{2}+\frac{2017}{3}+\frac{2017}{4}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)
\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)
\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
Giờ ta thế vào bài toán ban đầu được
\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)
\(=\frac{2017}{2018}\)
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Tính S = 1 + \(\frac{1}{2}\).(1+2) + \(\frac{1}{3}\).(1+2+3) + \(\frac{1}{4}\).(1+2+3+4) + .... + \(\frac{1}{2017}\).(1+2+3+....+2017)
\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{2017}.\left(1+2+3+...+2017\right)\)
\(S=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{2017}.\frac{\left(1+2017\right).2017}{2}\)
\(S=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2018}{2}\)
\(S=\frac{1}{2}.\left(2+3+4+...+2018\right)\)
\(S=\frac{1}{2}.\frac{\left(2+2018\right).2017}{2}\)
\(S=\frac{2020.2017}{4}=505.2017=1018585\)
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so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3+3\sqrt{4}}}+...+\frac{1}{2017\sqrt{2016}+2016\sqrt{2017}}\)
Tính giá trị của biểu thức .
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
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\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2016^2}+\frac{1}{2017^2}}\)
tính tổng
So sánh A và B nếu
\(A=\frac{-1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4}\)
\(B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)