Biết n! = 1.2.3. ... . n ( n \(\in\)N* )
Chứng tỏ rằng:
A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)
Biết n!=1.2.3.....n (n> hoặc =2)
chứng tỏ rằng:A=1/2!+2/3!+.......+2013/2014!<1
biết n! = 1.2.3....n (n thuộc N, n\(\ge\)2). chứng tỏ rằng A = \(\frac{1}{2}\)+\(\frac{2}{3}\)+ ....+ \(\frac{2013}{2014}\)< 1
Biết n!=1.2.3....n
CMR A=\(\frac{1}{2!}+\frac{2}{3!}+....+\frac{2013}{2014!}< 1\)
Biết n! = 1.2.3...n (Ví dụ: 3! = 1.2.3 = 6). chứng tỏ rằng S =\(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}< 2\)
Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)
\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)
\(\Rightarrow S< 2\) ( ĐPCM )
Chứng minh: \(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+..+\frac{n}{2^n}+...+\frac{2013}{2^{2013}}+\frac{2014}{2^{2014}}<2\)
Biết: n!= 1.2.3.....n (n\(\in\)N* ; n \(\ge\)2)
A = \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2013}{2014!}\)< 1
Biết n!=1.2.3...n \(\left(n\inℕ^∗;n\ge2\right)\)và \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+......+\frac{2014}{2015!}\)
Hãy so sánh A với 1
Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)
=> \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)
=> \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)
=> \(A=1-\frac{1}{2015!}< 1\)
Tìm n biết:
\(\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)=\frac{2013}{2014}\)
Với \(n\in\)N*
Cậu có thể vào đây tham khảo : http://h.vn/hoi-dap/question/119685.html
chịu thôi bạn ạ ko hiểu gì hết
So sánh P và Q biết : P = 2010/2011 + 2011/2012 + 2012/2013 và Q = 2010+2011+2012/ 2011 +2012+2013
Chứng tỏ N < 1 với N = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}+\frac{1}{2010^2}\)
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2010^2}