a*2+3=2013
A = (2013/2 + 2013/3+2013/4 + ....+2013/2014) : (2013/1+2012/2 +2011/3+...+1/2013)
\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(A=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(A=\frac{2013}{2014}\)
\(A=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2014}}{\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{1}{2013}\right)+1}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}}\)
\(=\frac{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}{2014.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(=\frac{2013}{2014}\)
Tính \(A=2013+\frac{2013}{1+2}+\frac{2013}{1+2+3}+\frac{2013}{1+2+3+4}+...+\frac{2013}{1+2+3+...+2012}\)
Ta có : 1 + 2 + 3 + ... + n = \(\frac{\left(n+1\right)n}{2}\)
Vậy nên : \(A=2013+\frac{2013}{\frac{3.2}{2}}+\frac{2013}{\frac{4.3}{2}}+...+\frac{2013}{\frac{2013.2012}{2}}\)
\(A=2013+\frac{4026}{2.3}+\frac{4016}{3.4}+...+\frac{4026}{2012.2013}\)
\(A=4026\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(A=4026\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=4026\left(1-\frac{1}{2013}\right)=4026.\frac{2012}{2013}=4024.\)
Cho A = 2013 + 20132 + 20133 + 20133 + 20134 + 20135 + 20136. Chứng tỏ rằng A chia hết cho 2
Ta có :
\(A=2013+2013^2+2013^3+2013^4+2013^5+2013^6\)
\(A=\left(2013+2013^2\right)+\left(2013^3+2013^4\right)+\left(2013^5+2013^6\right)\)
\(A=2013\left(1+2013\right)+2013^3\left(1+2013\right)+2013^5\left(1+2013\right)\)
\(A=2013.2014+2013^3.2014+2013^5.2014\)
\(A=2014\left(2013+2013^3+2013^5\right)\)
\(A=2.1007\left(2013+2013^3+2013^5\right)⋮2\)
\(\Rightarrow\)\(A⋮2\)
Vậy \(A⋮2\)
Chúc bạn học tốt ~
biết rằng tích A=(2013-1).(2013-2).(2013-3) ... (2013-n) có đúng 2013 thừa số. khi đó A = ?
Tích trên có 2013 thừa số suy ra n bằng 2013 => 2013 - 2013 = 0 =>A = 0
A = (2013-1) (2013-2) (2013-3)...(2013-n)
n là tổng vô giới hạn
Tuy nhưng trong đó có dãy 1;2;3;.....
Tồn tại số 2013
Vậy 2013 - 2013 = 0
Do đó Tích trên có kết quả là 0
Thực hiện tính :
a) A = 1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/2013(1+2+3+..+2013)
b) B = 1-3/7.3+2-4/2.4+3-5/3.5+4-6/4.6+....+2011-2013/2011.2013+2012-2014/2012.2014-2013+2014/2013.2014
Biết rằng tích A=(2013-1).(2013-2).(2013-3).....(2013-n) có đúng 2013 thừa số.Khi đó A bằng....
Có đúng 2013 thừa số
< = > Thừa số cuối cùng là 2013 - 2013 = 0
Tích là:
(2013-1).(2013-2).(2013-3).....0 = 0
Biết rằng tích A= (2013-1).(2013-2).(2013-3)...(2013-n) có đúng 2013 thừa số . Khi đó A bằng:
\(ChoA=\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+...+\frac{2013}{2013}\) và B=\(\frac{2013}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)Tính\(\frac{A}{B}\)
Tính nhanh
A = 2013 + \(\frac{2013}{1+2}\)+\(\frac{2013}{1+2+3}\)+\(\frac{2013}{1+2+3+4}\)+ ..........+\(\frac{2013}{1+2+3+....+2012}\)