Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
= 99x100x101
B = 99x100x101 : 3
= 333300
nhanh k minh
B= 1x2+3x4+5x6+...+99x100
=> Bx3= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ...+ 99x100x3
=> Bx3= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3)+...+99x100x(101-98)
=> Bx3= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +4x5x6 - 3x4x5 +...+ 99x100x101 - 98x99x100
=> Bx3= 99x100x101
=> B= 99x100x101:3
=> B= 333300
Gọi UCLN(2n + 3,3n + 4) là d
Ta có: 2n + 3 chia hết cho d => 3(2n + 3) chia hết cho d => 6n + 9 chia hết cho d
3n + 4 chia hết cho d => 2(3n + 4) chia hết cho d => 6n + 8 chia hết cho d
=> 6n + 9 - (6n + 8) chia hết cho d
=> 6n + 9 - 6n - 8 chia hết cho d
=> 1 chia hết cho d
=> d = 1
=> UCLN(2n + 3,3n + 4) = 1
Gọi d là ƯCLN (2n + 3 ; 3n + 4)
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\3n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+3\right)⋮d\\2\left(3n+4\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}6n+9⋮d\\6n+8⋮d\end{cases}}}\)
\(\Rightarrow6n+9-\left(6n+8\right)⋮d\)
\(6n+9-6n-8⋮d\)
\(1\) \(⋮d\)
\(\Rightarrow d=1\)
Vậy ƯCLN (2n + 3 ; 3n + 4) = 1
Bạn vào trang Wolfram Alpha sẽ thấy:
20182017 có 6667 chữ số
20172018 có 6669 chữ số
Vậy 20182017 < 20172018
1/1.3+1/3.5+1/5.7+.......+1/2003.2005
= 1/2.(2/1.3+2/3.5+2/5.7+.......+2/2003.2005)
= 1/2.(1 -1/3 + 1/3-1/5+1/5-1/7 + ...+ 1/2003 - 1/2005)
= 1/2.(1-1/2005)
= 1/2. 2004/2005
= 1002/2005
Ta có:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2004}\right)\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(\Rightarrow2\left(\frac{1}{1}-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2003}{2004}=\frac{2003}{4008}\)
Trả lời:
Ta có : 1.22= 1.2.2=1.2.(3-1)=1.2.3-1.2
2.32= 2.3.3=2.3.(4-1)=2.3.4-2.3
.................................................
98.992= 98.99.99=98.99.(100-1)=98.99.100-98.99
A=1.2.3 - 1.2 + 2.3.4 - 2.3 + ... + 98.99.100 - 98.99 hay A=1.2.3 + 2.3.4 +...+ 98.99.100 - (1.2 + 2.3 + ... + 98.99) = B - C
B=1.2.3 + 2.3.4 + ... + 98.99.100
B.4=1.2.3.4 + 2.3.4.(5 - 1) + ... + 98.99.100.(101 - 97)= 98.99.100.101
=> 98.99.100.101:4= 24497550
C=1.2 + 2.3 + ... + 98.99
C.3=1.2.3 + 2.3.(4 - 1) + ... + 98.99.(100 - 97)= 98.99.100
=> 98.99.100:3= 323400
Vậy A= 24497550 - 323400 = 24174150
B= 1.99+2.98+2.97+...98.2+99.1
=1.99+2.(99-1)+3.(99-2)+...+98.(99-97)+99.(99-98)
=1.99+2.99-1.2+3.99-2.3+...+98.99-97.98+99.99-98.99
=(1.99+2.99+3.99+...+98.99+99.99)-(1.2+2.3+3.4+...+97.98+98.99)
=99.(1+2+3+...+98+99)-(1.2+2.3+3.4+...+97.98+98.99)
=99.4950-(1.2+2.3+3.4+...+97.98+98.99)
=490050-(1.2+2.3+3.4+...+97.98+98.99)
Đặt C=1.2+2.3+3.4+...+97.98+98.99
=> 3C=1.2.3+2.3.3+3.4.3+...+97.98.3+98.99.3
=1.2.3+2.3.(4-1)+...+98.99.(100-97)
=1.2.3+2.3.4-1.2.3+...+98.99.100-97.98.99
=98.99.100
=> A=(98.99.100):3=323400
Vậy B=490050-323400=166650
=1.99+2.(99-1)+3.(99-2)+4.(99-3)+......+99.(99-98)
=99.(1+2+3+.......+99)-(2+2.3+3.4+........+98.99)
=99.(1+99).99:2-98.99.100:3
=99.50.99-98.33.100
=490050-323400=166650