Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.3+2.4+3.5+4.6+...+99.101
=(2-1).(2+1)+(3-1).(3+1)+(4-1).(4+1)+.....+(100-1).(100+1)
=2^2-1+3^2-1+4^2-1 +.....+100^2-1
=(2^2+3^2+4^2+.......100^2)-(1+1+....+1)
=(2^2+3^2+4^2+.......100^2)-99
\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)
A = 1(2+2)+2(3+2)+3(4+2)+.+99(100+2)
A = 1.2+1.2+2.3+2.2+3.4+3.2+.+99.100+99.2
A = (1.2+2.3+3.4+.+99.100)+2(1+2+3+.+99)
Phần còn lại bạn tự động não đi nha.
C= (4/1.3).(9/2.4).(16/3.5)...........(10000/99.101)
C=(4.9.16...........10000)/(1.3).(2.4)......(99.101)
C=(2^2.3^2.4^2..........100^2)/(1.2.3.4......99).(3.4.5.......101)
C=(2.3.4........100).(2.3.4......100)/(1.2.3.....99).(3.4.5....101)
Sau khi triệt tiêu ở tử và mẫu ta được:
C=(2.100)/101
C=200/101
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+.....+\frac{1}{8.10}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{7.9}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{8.10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{9}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{1}{2}\cdot\frac{8}{9}+\frac{1}{2}\cdot\frac{2}{5}=\frac{4}{9}+\frac{1}{5}=\frac{29}{45}\)
=1/2x(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.....+1/8-1/10)
=1/2x58/45
=29/45
Giải
Ta gọi T = (1^2+2^2+...+2005^2)-(1.3+2.4+3.5+...+2004.2006)
Đặt A = 1^2+2^2+3^2+...+2005^2
=> A = 1.1 + 2.2 +3.3 +...+ 2005.2005
=> A = 1.(2-1) + 2.(3-1) + 3.(4-1) +...+ 2005.(2006-1)
==> A = 1.2-1.1 + 2.3-1.2 + 3.4-1.3+...+2005.2006-1.2005
=> A = (1.2+2.3+3.4+...+2005.2006)-(1+2+3+...+2005)
Xét 1.2 +2.3+3.4+...+2005.2006
= 1/3.(1.2.3+2.3.3+...+2005.2006.3)
=1/3.[1.2.(3-0)+2.3.(4-1)+...+2005.2006.(2007-2004)]
=1/3.(1.2.3+2.3.4-1.2.3+...+2005.2006.2007-2004.2005.2006)
= 1/3 . 2005.2006.2007
= 2005.2006.2007/3 = 2690738070
Vậy A= 2690738070 - (1+3+5+...+2005)
=> A= 2690738070- [(2005-1):2+1].(2005+1)/2
=> A = 2690738070 - 1006009
=> A = 2689732061
Đắt B = 1.3+2.4+3.5+4.6+...+2003.2005 +2004.2006
=> B= (1.3+3.5+...+2003.2005)+(2.4+4.6+...+2004.2006)
=> 6B = (1.3.6+3.5.6+...+2003.2005.6)+(2.4.6+4.6.6+...+2004.2006.6)
=> 6B = [1.3.(5+1)+3.5.(7-1)+...+2003.2005.(2007-2001)] + [2.4.(6-0)+4.6.(8-2)+...+2004.2006.(2008-2002)]
=> 6B = (1.3.5+1.3.1+3.5.7-1.3.5+...+2003.2005.2007-2001.2003.2005)+(2.4.6+4.6.8-2.4.6+...+2004.2006.2008-2002.2004.2006)
=> 6B = 1.3.1+2003.2005.2007 + 2004.2006.2008
=> 6B = 16132350300
=> B = 16132350300/6 = 2688725050
Vì T = A - B = 2689732061-2688725050
=> T = 1007011