3x - (1/1x2+1/2x3+.....+1/99x100)=1/1x2x3+1/2x3x4+......+1/18x19x20
\(\Leftrightarrow3x-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+....+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\left(1-\frac{1}{100}\right)=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{1}{2}\cdot\frac{189}{380}\)
\(\Leftrightarrow3x-\frac{99}{100}=\frac{189}{760}\)
\(\Leftrightarrow3x=\frac{189}{760}+\frac{99}{100}=\frac{4707}{3800}\)
\(\Leftrightarrow x=\frac{1569}{3800}\)
\(\text{Vậy }x=\frac{1569}{3800}\)
Học sinh gương mẫu của lớp thầy Phú là đây
Tính :
a) 1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+.....+1
b)1/1x2x3+1/2x3x4+1/3x4x5+.....+1/98x99x100
bạn li-ke cho I love U thì ai giải cho bạn nữa
Tính :
a) 1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+.....+1
b)1/1x2x3+1/2x3x4+1/3x4x5+.....+1/98x99x100
C=1x2+2x3+3x4+........+99x100
D=1x2x3+2x3x4+3x4x5+.......+98x99x100
E=12+22+52+.......+992
F=1/1x2+1/2x3+..........+1/99x100
G=1/1x2x3+1/2x3x4+........+1/99x98x100
H= 1/1x2x3x4+1/3x4x5x6+............+1/97x98x99x100
K= 1+1/2(1+2)+1/3(1+2+3)+........+1/30(1+2+30)
L=1/21+1/22+1/23+1/24+..............+1/210
M=2015/2015x2017-20162
\(\frac{ }{ }\)
sao nhiều vậy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
F = 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Bài 1 Cho A= 10 mũ 2004 + 1/10 mũ 2005 + 1 và B = 10 mũ 2005 + 1/10 mũ 2006 + 1.So sánh A và B
Bài 2 Tính tổng các phân số sau
1/1x2 + 1/2x3 + 1/3x4 +...+1/20003x2004
1/1x3 + 1/3x5 + 1/3x7 +...+1/2019x2021
Bài 3 Hai can đựng 13 lít nước.Nếu bớt ở can thứ nhất 2 lít và thêm vào can thứ hai 9/2 lít,thì can thứ nhất nhiều hơn can thứ hai 1/2 lít.Hỏi lúc đầu mỗi can đựng được bao nhiêu lít nước
Tìm X biết : {44 + 2010/1x2 + 2006/2x3 + 2000/3x4 +........+ 32/44x45} = 44/45
A = 1x2 +2x3+3x4+........+99 x 100
B = 1x2x3 + 2x3x4 + 4x5x6 +...........+ 98 x99 x 100
tính A,B
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
A x 3 = 99x100x101
A = 99x100x101 : 3
A = 333300
Ta có:
\(A=1.2+2.3+3.4+...+99.100\)
\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)
\(\Leftrightarrow3A=99.100.101\Leftrightarrow A=\frac{99.100.101}{3}=333300\)
\(B=1.2.3+2.3.4+4.5.6+...+98.99.100\)
\(\Rightarrow4B=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+4.5.6.\left(7-3\right)+...+98.99.100.\left(101-97\right)\)
\(\Rightarrow4B=1.2.3.4+2.3.4.5-1.2.3.4+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100\)
\(\Leftrightarrow4B=98.99.100.101\Leftrightarrow B=\frac{98.99.100.101}{4}=24497550\)
A= 1 x 2 + 2 x 3 + 3 x 4 +........+ 99 x 100
=> 2 + 6 + 12 +........+ 9900
=> 8 + 12 +.....+ 9900
=> 20 +....+ 9900
=> 20 + 20 + 30 +....+ 9900
=> 70 +....+ 9900
=> ( 9900 x 70 ) : 2
=> 693000 : 2
=> 346500
B = 1 x 2 x 3 + 2 x 3 x 4 +......+ 98 x 99 x100
=> ( 1 x 2 x 3) + ( 2 x 3 x 4 ) +....+ ( 98 x 99 x 100 )
= 6 + 24 +.......+ 970200
=> 28 + 120 +...+ 970200
=> ( 148 x 970200 ) : 2
=> 143589600 : 2
=> 71794900
tinh
a)A=1x2+2x3+3x4+............+99x100
b)B=1x3+3x5+5x7+............97x99
c)C=1x2x3+2x3x4+..............98x99x100
1, a, Tính (2 cách)
A=1x2+2x3+3x4+....+nx(n+1)
b, Nêu cách tính tổng quát
c, áp dụng tính
B=1x2x3+2x3x4+....+(n-1)x(n+1)
giúp mình với đang cần gấp
a)
\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)
\(3A=1.2.3+2.3.3+3.4.3+...+n.\left(n+1\right).3\)
\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(3A=(1.2.3-0.1.2)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.5\right)+...+\left[n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\right]\)\(3A=-0.1.2+n.\left(n+1\right).\left(n+2\right)\)
\(3A=n.\left(n+1\right).\left(n+2\right)\)
\(A=\dfrac{n.\left(n+1\right).\left(n+2\right)}{3}\)
c)
\(B=1.2.3+2.3.4+...+\left(n-1\right).n.\left(n+1\right)\)
\(4B=1.2.3.4+2.3.4.4+3.4.5.4+...+\left(n-1\right).n.\left(n+2\right).4\)
\(4B=1.2.3.4+2.3.4.\left(5-1\right)+3.4.5.\left(6-2\right)+...+\left(n-1\right).n.\left(n+1\right).\left[\left(n+2\right)-\left(n-2\right)\right]\)\(4B=1.2.3.4+\left(2.3.4.5-1.2.3.4\right)+\left(3.4.5.6-2.3.4.5\right)+...+\left[\left(n-1\right).n.\left(n+1\right).\left(n+2\right)-\left(n-1\right).n.\left(n+1\right).\left(n-2\right)\right]\)\(4B=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\\ B=\dfrac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)