Tính S = 1 + 1/2.(1 + 2) + 1/3.(1 + 2 +3) + 1/4.(1 + 2 +3 + 4) +....+ 1/2017.(1 + 2 + 3 +...+ 2017)
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TÍnh
S=1+1/2(1+2)+1/3(1+2+3)+1/4+(1+2+3+4)+.....+1/2017(1+2+3+...+2017)
Tính tổng \(S=\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+...+\frac{2017}{2017^4+2017^2+1}\)
Tính tổng \(S=\dfrac{1}{1^4+1^2+1}+\dfrac{2}{2^4+2^2+1}+\dfrac{3}{3^4+3^2+1}+...+\dfrac{2017}{2017^4+2017^2+1}\)
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 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 = 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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tính m=2016+2016/2+2015/3+2014/4+...+1/2017/1/2+1/3+1/4+...+1/2017
Cho biểu thức : B = 2017+2017/1+2+2017/1+2+3+2017/1+2+3+4+....+2017/1+2+3+...+2012
1+1/2×(1+2)+1/3×(1+2+3)+1/4×(1+2+3+4)+....+1/2017×(1+2+3+....+2017)
( 1-1/2) . (1-1/3).(1-1/4).......(1-1/2016) . (1-1/2017)
=1/2.2/3.3.4x...x2015/2016.2016/2017
=1.2.3.4. ... .2015.2016/2.3.4.5. ... .2016.2017
(giống nhau you gạch đi )
=1/2017
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1+1/2×(1+2)+1/3×(1+2+3)+1/4×(1+2+3+4)+....+1/2017×(1+2+3+....+2017)
( 1-1/2) . (1-1/3).(1-1/4).......(1-1/2016) . (1-1/2017)
=1/2.2/3.3.4x...x2015/2016.2016/2017
=1.2.3.4. ... .2015.2016/2.3.4.5. ... .2016.2017
giống nhau you gạch đi cả hai bên số đó
=1/2017
nhé !
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1/1 nhân 1/2 +1/2 nhân 1/3 +1/3 nhân 1/4 +... +1/2016 nhân 1/2017
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