1/1.2.3 + 1/2.3.4 + 1/3.4.5 +............+1/2003.2004.2005 = ???????????????????
mong các bạn giúp đỡ
Mong mọi người giúp đỡ mình với:
1/2.3.4+1/3.4.5+1/4.5.6+...1/98.99.100
(1/1.2.3+1/2.3.4+1/3.4.5+1/4.5.6+1/5.6.7+...+1/8.9.10). x=23/45
các bạn giúp mình nhé , giải một cách thật nhanh mình mới tk
Ta có: \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
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=\(\frac{23}{11}\)
các bạn ơi gúp mk với
1/1.2.3+1/2.3.4+1/3.4.5+....+1/18.19.20
B= 1/ 1.2.3 + 1/ 2.3 4 + 1/ 3.4.5 + .... + 1/ 48.49.50
Mà ta có:
1/ 1.2 - 1/ 2.3 = 2/ 1.2.3
1/ 2.3 - 1/3.4 = 2/ 2.3.4
Từ đó=> B = 1/2 . ( 2/ 1.2.3 + 2/ 2,3.4 + ... + 2/ 18. 19. 20 )
= 1/2 .( 1/ 1.2 – 1/ 2.3 + 1/ 2.3 - .....- 1/19.20)
= 1/2. ( 1/ 1.2 – 1/ 19.20 ) = 1/ 2 . 189/380 = 189/760
P = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/n(n+1)(n+2)
S = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/48.49.50 .
tao có:
2p=2/1.2.3+2/2.3.4+...+2/n.n(+1)n(n+2)
2p=3-1/1.2.3+4-2/1.2.3+...+(n+2)-n/n.(n+1).(n+2)
2p=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+(n+2)/n.(n+1).(n+2)-n/n.(n+1).(n+2)
2p=1/1.2-1/2.3+1/2.3-1/3.4+...+1/n.(n+1)-1/(n+1).(n+2)
2p=1/1.2-1/(n+1).(n+2)
2p=(n+!).(n+2)-2/(2n+2).(n+2)
suy ra p=(n+1).(n+2)-2/(2n+2).(2n+4)
2s=3-1/1.2.3+4-2/1.2.3+...+50-48/48.49.50
2s=3/1.2.3-1/1.2.3+4/2.3.4-2/2.3.4+...+50/49.50.48-48/48.50.49
2s=1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
2s=1/1.2-1/49.50
'2s=1/2-1/2450
2s=1225/2450-1/2450
2s=1224/2450
s=612/1225
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)1
\(P=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(P=\frac{\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)}{2}\)
S cx tinh giong v
Bài 4:
a) Chứng minh các công thức sau:
A = 1.2.3+2.3.4+3.4.5+...+(n-2)(n-1)n = (n−2).(n−1).n.(n+1):
4
b) Áp dụng tính tổng sau: G = 1.2.3 + 2.3.4 + 3.4.5 +...+ 2021.2022.2023
4A = 4.[1.2.3 + 2.3.4 + 3.4.5 + … + (n – 1).n.(n + 1)]
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + … + (n – 1).n.(n + 1).4
4A = 1.2.3.4 + 2.3.4.(5 – 1) + 3.4.5.(6 – 2) + … + (n – 1).n.(n + 1).[(n + 2) – (n – 2)]
4A = 1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)
4A = (n – 1).n(n + 1).(n + 2)
A = (n – 1).n(n + 1).(n + 2) : 4.
cau a thi sao ha ban ?
ok thanks ban nhe
Giải giúp m bài này.
1/1.2.3 + 1/2.3.4 + 1/3.4.5 +......+ 1/998.999.1000
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\left(\frac{499500}{999000}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Cho S= 1.2.3 + 2.3.4 + 3.4.5 + ... + k(k+1)(k+2)
Chứng minh rằng 4S + 1 là số chính phương.
Cho e hỏi là vì sao khi :
S.4=1.2.3.4+2.3.4.4+...+k(k+1)(k+1).4
=1.2.3(4-0)+2.3.4.(5-1)+...+k(k+1)(k+2)(k+3-k-1)
Tới đoạn này thì S lại bằng:
=1.2.3.4-0+1.2.3.4-2.3.4.5+...+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
Và sau đó chỉ còn: =(k-1)k(k+1)(k+2)
MONG CÁC BẠN, CÁC THẦY CÔ GIẢI ĐÁP GIÚP MÌNH!!!
a) b=1/3+1/15+1/35+...+1/97.99
b) c=2/1.2.3+2/2.3.4+2/3.4.5+...+2/98.99.100
c) d=5/2.3.4+5/3.4.5+...+5/98.99.100+5/99.100.101
GIẢI GIÚP MÌNH THEO CÁCH HỌC CỦA LỚP 6 VỚI Ạ. CẢM ƠN MỌI NGƯỜI NHIỀU!
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/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
Thực hiện phép tính:
1/1.2.3 + 1/2.3.4 + 1/3.4.5 +......+ 1/2007.2008.2009
làm ơn giúp tớ với !!!!!!!!
1/1.2.3+1/2.3.4+...+1/2007.2008.2009=1-1/2-1/3+1/2-1/3-1/4+...-1/2008-1/2009=1-1/2009=2008/2009
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{2007.2008.2009}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-.....-\frac{1}{2008.2009}\)
\(=\frac{1}{1.2}-\frac{1}{2008.2009}=\frac{1}{2}-\frac{1}{4034072}=\frac{2017035}{4034072}\)