CMR:\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Những câu hỏi liên quan
CMR:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
a) Cho tổng gồm 2014 số hạng
S= \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\)
CMR S<1
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+....+\frac{2014}{4^{2014}}\)
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\)
\(4S-S=\left(1+\frac{2}{4}+\frac{3}{4^2}+\frac{4}{4^3}+...+\frac{2014}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+\frac{4}{4^4}+...+\frac{2014}{4^{2014}}\right)\)
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
\(12S=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\)
\(12S-3S=\left(4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}-\frac{2014}{4^{2013}}\right)-\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\right)\)
\(9S=4-\frac{2014}{4^{2013}}-\frac{1}{4^{2013}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{4028}{4^{2014}}-\frac{4}{4^{2014}}+\frac{2014}{4^{2014}}\)
\(9S=4-\frac{2010}{4^{2014}}< 4\)
\(\Rightarrow9S< 4\)
\(\Rightarrow S< \frac{4}{9}< 1\)(đpcm)
Đúng 0
Bình luận (0)
Ta có :
\(S=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)( 1 )
\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)( 2 )
Lấy ( 2 ) - ( 1 ) ta được :
\(3S=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)
gọi \(B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)( 3 )
\(4B=4+1+\frac{1}{4}+...+\frac{1}{4^{2012}}\) ( 4 )
Lấy ( 4 ) - ( 3 ) ta được :
\(3B=4-\frac{1}{4^{2013}}\)
\(\Rightarrow B=\frac{4-\frac{1}{4^{2013}}}{3}=\frac{4}{3}-\frac{1}{4^{2013}.3}\)
\(\Rightarrow3S=\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}\)
\(\Rightarrow S=\frac{\frac{4}{3}-\frac{1}{4^{2013}.3}-\frac{2014}{4^{2014}}}{3}=\frac{4}{9}-\frac{1}{4^{2013}.9}-\frac{2014}{4^{2014}.3}< \frac{4}{9}< 1\)
vậy \(S< 1\)
Đúng 0
Bình luận (0)
\(S=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2014}{2^{2013}}+\frac{2015}{2^{2014}}\)cmr S<4
cmr S<1/2 khi S = \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)
Cho tổng gồm 2014 số hạng: S= \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\).CMR: S<\(\frac{1}{2}\)
Rút gọn biểu thức : \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+\frac{3}{3^4+3^2+3}+...+\frac{2014}{2014^4+2014^2+2014}\)
Chứng minh: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Đây là bài chứng minh chứ ko phải tính đúng ko?
Đúng 0
Bình luận (0)
Tính:
A= 2014 + \(\frac{2014}{1+2}+\frac{2014}{1+2+3}+\frac{2014}{1+2+3+4}+........+\frac{2014}{1+2+3+4+.....+2013}\)
\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
Đúng 0
Bình luận (0)
Chứng minh:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)