1/1.2.3+1/2.3.4+1/3.4.5+....+1/49.50.51
Giải nhanh giúp mình với , cái dấu / là vạch giữa tử và mẫu
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Những câu hỏi liên quan
HL
2
7 tháng 7 2016
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{1999.2000}-\frac{1}{2000.2001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2000.2001}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4002000}\right)=\frac{1}{2}\left(\frac{2000999}{4002000}\right)=\frac{2000999}{8004000}\)
7 tháng 7 2016
A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/1999.2000.2001
A = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + 2/3.4.5 + ... + 2/1999.2000.2001)
A = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/1999.2000 - 1/2000.2001)
A = 1/2.(1/1.2 - 1/2000.2001)
A = 1/2.(1/2 - 1/4002000)
Đến đây số to wa, bn tự lm típ
Chú ý: tính hiệu giữa: 1/1.2 - 1/2.3 = 3/1.2.3 - 1/1.2.3 = 2/1.2.3, nhân thêm 2 vào tử
Ủng hộ mk nha ^_-
NA
5
N
12 tháng 10 2018
đặt S=1.2.3+2.3.4+....+47.48.49
4S=1.2.3.(4-0)+2.3.4.(5-1)+...+47.48.49.(50-46)
4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+....+47.48.49.50-46.47.48.49
4S=47.48.49.50-1.2.3
S=(47.48.49.50-1.2.3):4
AK
15 tháng 4 2018
\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2009.2010.2011}\)
\(S=2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2009.2010.2011}\right)\)
\(S=2.\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2009.2010}-\frac{1}{2010.2011}\right)\)
\(S=1.\left(\frac{1}{1.2}-\frac{1}{2010.2011}\right)\)
\(S=\frac{1}{1.2}-\frac{1}{2010.2011}\)
\(S=\frac{1}{2}-\frac{1}{2010.2011}< \frac{1}{2}\)
Vậy \(S< \frac{1}{2}\)
Chúc bạn học tốt !!!
AK
15 tháng 4 2018
Áp dụng công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
Chúc bạn học tốt !!!
S
23 tháng 2 2017
Ta có: \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
23 tháng 2 2017
Đặt A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}\) \(+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{9.10}\)
2A=\(\frac{22}{45}\)
A=\(\frac{22}{45}\div2\)
A=\(\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy x=\(\frac{23}{11}\)
28 tháng 4 2019
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\frac{370}{741}\)
\(A=\frac{185}{741}\)
H
28 tháng 4 2019
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
Tự tính tiếp nha =)) mỏi tay quá
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 7 2023
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
3
26 tháng 9 2021
Ta có \(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
Áp dụng:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{10\cdot11\cdot12}\\ =\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}-\dfrac{1}{11\cdot12}\\ =\dfrac{1}{2}-\dfrac{1}{11\cdot12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)
26 tháng 9 2021
sai rồi kìa
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\) mà
29 tháng 11 2016
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 28.29.30
4A = 1.2.3.(4-0) + 2.3.4.(5-1) + 3.4.5.(6-2) + ... + 28.29.30.(31-27)
4A = 1.2.3.4 - 0.1.2.3. + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 28.29.30.31 - 27.28.29.30
4A = 28.29.30.31 - 0.1.2.3
4A = 28.29.30.31
\(A=\frac{28.29.30.31}{4}=7.29.30.31=188790\)
Theo cách tính trên ta dễ dàng tính được:
1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n + 1) = \(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{49.50.51}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-....-\frac{1}{50.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2550}\right)=\frac{637}{2550}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\)
ta có dạng tổng quát
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)-\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) bạn quy đồng ra rồi tính nha
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{49.50}-\frac{1}{50.51}\)
\(2A=\frac{1}{1.2}-\frac{1}{50.51}\)
\(2A=\frac{637}{1275}\)
\(A=\frac{637}{2550}\)