K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
12 tháng 10 2016
A=1.2.3+2.3.4+4.5.6+___+19.20.21
4A=1.2.3.4+2.3.4.4+3.4.5.4+___+19.20.21.4
=1.2.3.(4-0)+2.3.4(5-1)+3.4.5(6-2)+___+19.20.21.(22-18)
=1.2.3.4-0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+___+19.20.21.22-19.20.21.18
=(1.2.3.4-1.2.3.4)+(2.3.4.5-2.3.4.5)+___+(19.20.21.18-19.20.21.18)+19.20.21.22
A=19.20.21.22:4
A =43 890
27 tháng 4 2017
A= \(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+ ... + \(\frac{1}{19.20.21}\)< \(\frac{1}{4}\)
= 1 - \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{19}-\frac{1}{20}-\frac{1}{21}\)
= 1 - ( \(\frac{1}{2}-\frac{1}{3}\)+ \(\frac{1}{2}-\frac{1}{3}\)) + ... + ( \(\frac{1}{19}-\frac{1}{20}+\frac{1}{19}-\frac{1}{20}\)) - \(\frac{1}{21}\)
= 1 - \(\frac{1}{21}\)
= \(\frac{20}{21}\)< \(\frac{1}{4}\)
=> Đề bài có sai ko bạn?
U
9 tháng 8 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\right)\)
\(2B=1-\frac{1}{3^{21}}\)
\(B=\frac{1-\frac{1}{3^{21}}}{2}\)
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(C=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)\)
\(C=\frac{1}{2}\cdot\frac{209}{420}\)
\(C=\frac{209}{480}\)
DP
3
DD
6 tháng 8 2015
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{19.20.21}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
LC
6 tháng 8 2015
=\(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{21-19}{19.20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{1}{20.21}\right)\)
=\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{420}\right)=\frac{1}{2}.\frac{209}{420}=\frac{209}{840}\)
TQ
1
MC
6 tháng 5 2016
\(A=\frac{24}{1.2.3}+\frac{24}{2.3.4}+....+\frac{24}{19.20.21}\)
\(A=24.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{19.20.21}\right)\)
\(A=12.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{19.20.21}\right)\)
\(A=12.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-....-\frac{1}{20.21}\right)\)
\(A=12.\left(\frac{1}{2}-\frac{1}{420}\right)=12.\frac{209}{420}=\frac{209}{35}\)
LB
7 tháng 5 2018
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
8 tháng 5 2018
\(P=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)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
=30076