So sánh tổng gồm 1006 số hạng :
\(S=\dfrac{1}{1\cdot1\cdot3}+\dfrac{1}{2\cdot3\cdot5}+...+\dfrac{1}{1006\cdot2011\cdot2013}\)
K
Khách
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Những câu hỏi liên quan
TA
21 tháng 7 2021
`A=1/(1.3)+1/(3.5)+....+1/(2009.2011)`
`=> 2A=2/(1.3)+1/(3.5)+...+2/(2009.2011)`
`=1-1/3+1/3+1/5+.....+1/2009-1/2011`
`=1-1/2011`
`=2010/2011`
`=> A=1005/2011`
H
21 tháng 7 2021
Dựa trên công thức: \(\dfrac{a}{n.\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\), ta có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
=\(\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{2010}{2011}\right)\)
= \(\dfrac{2010}{4022}\)
= tự rút gọn nhé
Hok tốt!
L
25 tháng 3 2018
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)
\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)
6 tháng 8 2018
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
17 tháng 4 2017
Đặt :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{2010.2011}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{2010}-\dfrac{1}{2011}\)
\(A=1-\dfrac{1}{2011}\)
\(A=\dfrac{2010}{2011}\)
~ Chúc bn học tốt ~
17 tháng 4 2017
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2010}-\dfrac{1}{2011}\)
\(=\dfrac{1}{1}-\dfrac{1}{2011}\)
\(=\dfrac{2010}{2011}\)
27 tháng 12 2018
Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
=>2A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+...+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\)\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)
=>A=\(\dfrac{n^2+3n}{4n^2+12n+8}\)
