CMR : 1/3-2/3^2+3/3^3-4/3^4+...-2014/3^2014<1/5
bai 1) tim x, y
x.y-x+2y=3
bai2 ) cmr
a)1/2-1/4+1/8-1/16+1/32-1/64<1/3
b) 1/3-2/3^2+3/3^3-4/3^4+....+99/3^99-100/3^100<3/16
c)1cho tổng gồm 2014 số hạng : s=1/4+2/4^2+3/4^3+.....2014/4^2014
CMR:\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)
Bài 1 : Tính tổng
a) 1 *2 *3 + 2 * 3 *4 + 3 * 4 * 5 + ... + 2013 * 2014 * 2015 + 2014 * 2015 * 2016
b) 1 * + 3 * 4 + 5 * 6 + ... + 99 * 100
Bài 2 : CMR : 1^3 + 2^3 + 3^3 + ... + n^3 = ( 1 + 2 + 3 + ... + n )^2
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)
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 \(S< 1\)
cmr S<1/2 khi S = \(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\)
\(S=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2014}{2^{2013}}+\frac{2015}{2^{2014}}\)cmr S<4
CMR : 1/2!+2/3!+3/4! + ...+2013/2014! < 1
ủa, nó nhỏ hơn 1 mà sao bn lại ghi lớn hơn 1
\(\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2013}{1.2...2014}\)
\(=\frac{1}{2}+\frac{1}{1.3}+\frac{1}{1.2.4}+...+\frac{1}{1.2...2012.2014}\)
\(=\frac{1.1.3.4...2012.2014}{2.1.3.4...2012.2014}+\frac{1.2.4.5...2012.2014}{1.3.2.4.5...2012.2014}+...+\frac{1}{1.2.....2012.2014}\)(Quy đồng mẫu)
\(=\frac{1.1.3.4...2012.2014+1.2.4.5...2012.2014+...+1}{1.2...2012.2014}>1\)
Tính tổng:
A= 1+2014^1+2014^2+2014^3+...+2014^2014+2014^2015
B = 3-3^2+3^3+3^4+...+3^100
A = 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015
2014A = 2014^1 + 2014^2 + 2014^3 + 2014^4 + ... 2014^2015 + 2014^2016
2014A - A = ( 2014^1 + 2014^2 + 2014^3 + 2014^4 + .... + 2014^2015 + 2014^2016 ) - ( 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015 )
2013A = 2014^2016 - 1
A = 2014^2016 - 1 / 2013
B = 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 ( đề hơi vui )
3B = 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101
3B - B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - ( 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 )
2B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - 3 + 3^2 - 3^3 - 3^4 - ... - 3^100
2B = 3^2 - 3^3 + 3^101 - 3 + 3^2 - 3^3
2B = 9 - 27 + 3^101 - 3 + 9 - 27
2B = -18 + 3^101 - 3 + ( -18 )
2B = -39 + 3^101
B = -39 + 3^101 / 2
A = 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015
2014A = 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016
2014A - A = ( 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016 ) - ( 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015 )
2013A = 20142016 - 1
A \(=\frac{2014^{2016}-1}{2013}\)
B = 3 - 32 + 33 - 34 + ... + 3100
3B= 32 - 33 + 34 - 35 + ... + 3101
3B + B = ( 3 - 32 + 33 - 34 + ... + 3100 ) + ( 32 - 33 + 34 - 35 + ... + 3101 )
4B = 3 + 3101
B = \(\frac{3+3^{101}}{4}\)