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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
A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 20.21.22
4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + 20.21.22.4
4A = 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) + ... + 20.21.22.(23 - 19)
4A = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 20.21.22.23 - 19.20.21.22
4A = 20.21.22.23
A = 20.21.22.23 : 4
A = 53130
Đặt A = 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 +...+ 20.21.22
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+4.5.6.4+...+20.21.22.4\)
\(=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+4.5.6.\left(7-3\right)+...+20.21.22.\left(23-19\right)\)
= 1.2.3.4 + 2.3.4.5 -1.2.3.4 + 3.4.5.6 - 2.3.4.5 + 4.5.6.7 - 3.4.5.6 +...+ 20.21.22.23 - 19.20.21.22
= 20.21.22.23
= 212520
_Hok tốt_
!!!
Mình nghĩ đề bài thế này mới đúng : 1.2.3+2.3.4+...+48.49.50
Đặt B = 1.2.3 + 2.3.4 + ....... + 48.49.50
B = 1.2.3 + 2.3.4 +....+ 48.49.50
\(\Rightarrow\)4B= 1.2.3.4 + 2.3.4.4 +.....+48.49.50.4
=1.2.3.4 + 2.3.4.(5-1) +.....+48.49.50.(51-47)
=1.2.3.4 + 2.3.4.5 - 1.2.3.4 +.....+ 48.49.50.51 - 47.48.49.50
=48.49.50.51
= 1499400
Vậy 1.2.3 + 2.3.4 + ....... + 48.49.50 = 1499400
Gọi d là \(ƯC\left(2n+3;3n+4\right)\)
Ta có: \(2n+3⋮d\Rightarrow3\left(2n+3\right)⋮d\Leftrightarrow6n+9⋮d\)
\(3n+4⋮d\Rightarrow2\left(3n+4\right)⋮d\Rightarrow6n+8⋮d\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\)
\(\Rightarrow6n+9-6n-8⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d\inƯ\left(1\right)=\left\{1\right\}\)
Vậy \(ƯCLN\left(2n+3;3n+4\right)=1\left(đpcm\right)\)
B=1.2.3+2.3.4+4.5.6+....+20.21.22
=>4B=4(1.2.3+2.3.4+3.4.5+...+20.21.22)
=>4B=1.2.3.4+2.3.4.4+3.4.5.4+...+20.21.22.4
=>4B=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+20.21.22(23-19)
=>4B=1.2.3.4+2.3.4.5-2.3.4+3.4.5.6-2.3.4.5+...+20.21.22.23-19.20.21.22
=>4B=20.21.22.23
=>4B=212520
=>B=53130
mình nha
Đặt A = 1 . 2 . 3 + 2 . 3 . 4 + 3 . 4 . 5 + 4 . 5 . 6 + ... + 20 . 21 . 22
=> 4A = 1 . 2 . 3 . 4 + 2 . 3 . 4 . 4 + 3 . 4 . 5 . 4+ 4 . 5 . 6 . 4 + ... + 20 . 21 . 22 . 4
=> 4A = 1 . 2 . 3 . ( 4 - 0 ) + 2 . 3 . 4 . ( 5 - 1 ) + 3 . 4 . 5 . ( 6 - 2 ) + 4 . 5 . 6 . ( 7 - 3 ) + ... + 20 . 21 . 22 . ( 23 - 19 )
=> 4A = 1 . 2 . 3 . 4 - 0 + 2 . 3 . 4 . 5 - 1 . 2 . 3 . 4 + 3 .4 . 5 . 6 - 2 . 3 . 4 . 5 + 4 . 5 . 6 . 7 - 3 . 4 . 5 . 6 + ... + 20 . 21.22.23 - 19.20.21.22
=> 4A = 0 + 20 . 21 . 22 . 23
=> 4A = 212520
=> A = \(\frac{212520}{4}=53130\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{20.21.22}=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{20.21.22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{462}\right)=\frac{1}{2}.\frac{115}{231}=\frac{115}{462}\)
Ta có :
\(M=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{10.11.12}\)
\(M=5.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\right)\)
\(M=5.\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{5}{2}.\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
\(M=\frac{5}{2}.\left(\frac{66}{132}-\frac{1}{132}\right)\)
\(M=\frac{5}{2}.\frac{65}{132}\)
\(M=\frac{325}{264}\)
Tham khảo nha !!! Chúc học tốt !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{1}{3}=\frac{1}{1.2.3}\)
Đặt A=1.2.3+2.3.4+.....+5.6.7
4A=1.2.3.4+2.3.4.(5-1)+.........+5.6.7.(8-4)
4A=1.2.3.4+2.3.4.5-1.2.3.4+........+5.6.7.8-4.5.6.7
4A=5.6.7.8
A=5.6.7.8:4
A=420