A = 1.2.3 + 2.3.4 + 3.4.5 + ......................+ 17.18.19
tính ???? tau ko hỏi mấy đứa lớp 6c cxh
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/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
Tính B/A biết:
A=1/1.2+1/2.3+...+1/n(n+1)+...+1/2008.2009
B=1/1.2.3+1/2.3.4+1/3.4.5+...+1/n.(n+1)(n+2)+1/2008.2009.2010
Đây là dạng lớp 7 đúng ko mọi người
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}=\frac{1}{1}+\left(-\frac{1}{2}+\frac{1}{2}\right)+...+\left(-\frac{1}{2008}+\frac{1}{2008}\right)-\frac{1}{2009}\)
\(A=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2.B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2008.2009.2010}\)
\(2.B=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2008.2009}-\frac{1}{2009.2010}\right)\)
\(2.B=\frac{1}{1.2}+\left(-\frac{1}{2.3}+\frac{1}{2.3}\right)+...+\left(-\frac{1}{2008.2009}+\frac{1}{2008.2009}\right)-\frac{1}{2009.2010}\)
\(2.B=\frac{1}{1.2}-\frac{1}{2009.2010}=\frac{2009.2010-1.2}{2009.2010}\)
=> \(B=\frac{2009.1005-1}{2009.2010}\)
Vậy \(\frac{B}{A}=\frac{2009.1005-1}{2009.2010}:\frac{2008}{2009}=\frac{2009.1005-1}{2008.2010}=...\)
A= 1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100
A= 1.2.3 + 2.3.4 + 3.4.5 +.....+ 98.99.100
4A = 98.99.100.4 + .....+ 3.4.5.4 + 2.3.4.4 + 1.2.3.4
4A = 98.99.100.(101-97) +... + 2.3.4.(5-1) + 1.2.3.4
4A = 98.99.100.101 - 97.98.99.100+......+2.3.4.5 - 1.2.3.4 + 1.2.3.4
4A = 98.99.100.101
A = 98.99.100.101 : 4
A = 24497550
Tinh nhanh
A= \(1.2.3+2.3.4+3.4.5+...+48.49.50\)
B = \(1.2.3+2.3.4+3.4.5+...+n.\left(n+1\right).\left(n+2\right)\)
A = 1.2.3 + 2.3.4 + ....+ 48.49.50
=> 4A = 1.2.3.4 + 2.3.4.(5-1) + ...+ 48.49.50.(51-17)
= 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + .....+ 48.49.50.51 - 47.48.49.50
= 48.49.50.51
=> A = 48.49.50.51:4 = 12.49.50.51
bài b) làm tương tự nha
Tính : a, 1.2.3 + 2.3.4 + 3.4.5 + ... + (n - 1).n.(n+1)
b, 1.2.3 + 3.4.5 + 5.6.7 + 98.99.100
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254
549 ,1326 ở đâu zậy bạn !!! :/
A= 1.2.3+2.3.4+3.4.5+..........+48.49.50
\(A=1.2.3+2.3.4+3.4.5+...+48.49.50\)
\(4A=1.2.3.4+2.3.4.4+3.4.5.4+...+48.49.50.4\)
\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+48.49.50.\left(51-47\right)\)
\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+48.49.50.51-48.48.49.50\)
\(4A=48.49.50.51\)
\(A=\dfrac{48.49.50.51}{4}=1499400\)
4A=4.[1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + . . . . . . + 48.49.50]
4A=1.2.3.4 +2.3.4.4 +3.4.5.4 +4.5.6.4 +.........+48.49.50.4
4A=1.2.3.4 +2.3.4.(5-1) + 3.4.5.(6-2) +4.5.6(7-3)+....+ 48.49.50(51-47)
4A=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+....+ 48.49.50.51--47.48.49.50
4A =48.49.50.51
A=(48.49.50.51)/4
A=1.2.3+2.3.4+3.4.5+......+88.89.90
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+88.89.90.4
4A=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+...+88.89.90.(91-87)
4A=1.2.3.4+1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+...+88.89.90.91-87.88.89.90
4A=88.89.90.91
A=16036020
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
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