\(\text{Tính tổng: }\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
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16 tháng 3 2017
Đặt \(A=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
Ta có:
\(3A=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{6}-\dfrac{1}{24360}\)
\(\Rightarrow3A=\dfrac{1353}{8120}\)
\(\Rightarrow A=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(A=\dfrac{451}{8120}\)
KT
1
BT
12 tháng 3 2018
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)
=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)
=> \(A=\frac{451}{8120}\)
LN
13
10 tháng 5 2015
Nhận xét: 1/1.2.3 - 1/2.3.4 = 3/1.2.3.4, 1/2.3.4 - 1/3.4.5 =3/2.3.4.5,...,1/27.28.29 - 1/28.29.30
Gọi biểu thức phải tính bằng A,ta tính được:
3A=1/2.3 - 1/28.29.30 = 4059/28.29.30
vậy A = 1353/8120
HN
9 tháng 10 2017
\(\dfrac{B}{2}=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{2014.2015.2016.2017}\)
\(\Leftrightarrow\dfrac{3B}{2}=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{2014.2015.2016.2017}\)
\(=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2014.2015.2016}-\dfrac{1}{2015.2016.2017}\)
\(=\dfrac{1}{1.2.3}-\dfrac{1}{2015.2016.2017}\)
Tự làm nốt nhé
3 tháng 7 2018
\(a.C=\dfrac{x^4+x^8+x^{12}+x^{16}+x^{20}+x^{24}+x^{28}+1}{x^3+x^7+x^{11}+x^{15}+x^{19}+x^{23}+x^{27}+x^{31}}=\dfrac{x^{28}+x^{24}+...+x^8+x^4+1}{x^3\left(x^{28}+x^{24}+...+x^8+x^4+1\right)}=\dfrac{1}{x^3}\) Tại x = 2015 thì : \(C=\dfrac{1}{x^3}=\dfrac{1}{2015^3}\)
\(b.F=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{2011.2012.2013.2014}\)
\(3F=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{2011.2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{3.4.5}-\dfrac{1}{4.5.6}+...+\dfrac{1}{2011.2012.2013}-\dfrac{1}{2012.2013.2014}\)
\(3F=\dfrac{1}{1.2.3}-\dfrac{1}{2012.2013.2014}\)
Tới đây dễ rồi , bạn tự tính nốt .
3 tháng 7 2018
Vì làm vậy để triệt tiêu dần mà ( dang bài kiểu ... này thường là phải triệt tiêu ) Triệu Tử Dương
KT
1
30 tháng 3 2016
a,\(\frac{2}{3.5}+\frac{2}{5.7}+.......+\frac{2}{11.13}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.............+\frac{1}{11}-\frac{1}{13}\)
=\(\frac{1}{3}-\frac{1}{13}\)
=\(\frac{10}{39}\)
b,Đặt A=\(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.............+\frac{1}{27.28.29.30}\)
3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...........+\frac{3}{27.28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+.............+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
3A=\(\frac{1}{6}-\frac{1}{24360}\)
3A=\(\frac{1353}{8120}\)
A=\(\frac{451}{8120}\)
27 tháng 3 2017
\(c=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(c=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)\(=\)\(\dfrac{1}{6}-\dfrac{1}{24360}\)
\(C=\) \(\dfrac{4059}{24360}\)
27 tháng 3 2017
Ta có:
\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1353}{8120}\)
\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(C=\dfrac{451}{8120}\)
Ta có \(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\dfrac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29\cdot30}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot2\cdot3\cdot4}+\dfrac{3}{2\cdot3\cdot4\cdot5}+...+\dfrac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{2\cdot3\cdot4}-\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{27\cdot28\cdot29}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{1\cdot2\cdot3}-\dfrac{1}{28\cdot29\cdot30}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{6}-\dfrac{1}{24360}\right)=\dfrac{1}{3}\cdot\dfrac{1353}{8120}=\dfrac{451}{8120}\)
\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{27.28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\right)=\dfrac{1}{3}.\dfrac{4060-1}{28.29.30}\)
\(=\dfrac{1}{3}.\dfrac{4059}{24360}=\dfrac{1353}{24360}=\dfrac{451}{8120}\)