b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6

=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52

=48*50*52

=>B=20800

d: 9D=1*4*9+4*7*9+...+46*49*9

=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43

=1*2*4+46*49*52

=117216

=>D=13024

a: 

Cách khác của bài 1:

B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100

B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)

B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98

B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)

B=98.99.1003+98.992B=98.99.1003+98.992

B=323400+4851=328251B=323400+4851=328251 

1.3+2.4+3.5+...+98.100=22−1+32−1+...+992−1=12+22+32+...+992−99=99.100.1996−99=3282511.3+2.4+3.5+...+98.100=22−1+32−1+...+992−1=12+22+32+...+992−99=99.100.1996−99=328251
Bài 2: A=1.2.3+2.3.4+...+97.98.99<=>4A=1.2.3.4+2.3.4.4+...+97.98.99.4=1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)A=1.2.3+2.3.4+...+97.98.99<=>4A=1.2.3.4+2.3.4.4+...+97.98.99.4=1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)
1.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+97.98.99.100−96.96.98.99=97.98.99.1001.2.3.(4−0)+2.3.4.(5−1)+...+97.98.99.(100−96)=1.2.3.4−0.1.2.3+2.3.4.5−1.2.3.4+...+97.98.99.100−96.96.98.99=97.98.99.100
Suy ra A=97.98.99.1004=23527350A=97.98.99.1004=23527350 

đc rùi hì hì

thanks nha♥

Đợi đứa khác nhé!☻

10140,22

k cho minh nha

1.3+2.4+3.5+...+99.101

=1.(2+1)+2.(3+1)+3.(4+1)+...+99.(100+1)

=1.2+1+2.3+2+3.4+3+...+99.100+99

=(1.2+2.3+3.4+...+99.100)+(1+2+3+...+99)

Đặt A=1.2+2.3+...+99.100

=>3A=1.2.3+2.3.3+...+99.100.3

=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)

=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100

=>3A=99.100.101=999900=>A=333300

Đặt B=1+2+3+...+99

Số số hạng của B là (99-1).1+1=99

=>(99+1).99:2=4950

Mà lại có:1.3+2.4+3.5+...+99.101=A+B=333300+4950=338250

ak tìm tổng của biểu thức đấy bn ak

\(P=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{2021.2023}\)

Ta sẽ "tách" P làm 2 phần:

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)

\(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)

Do đó \(P=A+B\)

Ta có \(A=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{2023}\right)\) 

\(A=\dfrac{1011}{2023}\)

Mặt khác, \(B=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2020.2022}\)

\(B=\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2020.2022}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+\dfrac{8-6}{6.8}+...+\dfrac{2022-2020}{2020.2022}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)

\(B=\dfrac{505}{2022}\)

Từ đó \(P=A+B=\dfrac{1011}{2023}+\dfrac{505}{2022}=\dfrac{3065857}{4090506}\)

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