=2 

day la do meo

Answer:

Chứng tỏ không phải số nguyên nhỉ?

\(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)

\(\Rightarrow A.\frac{3}{4}=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2010}+\left(\frac{3}{4}\right)^{2011}\)

\(\Rightarrow\frac{3}{4}A+A=\left(\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2010}+\left(\frac{3}{4}\right)^{2011}\right)+\left(1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\right)\)

\(\Rightarrow\frac{7}{4}A=\left(\frac{3}{4}\right)^{2011}+1\)

\(\Rightarrow A=\frac{4.\left(\frac{3}{4}\right)^{2011}+4}{7}\)

Vậy A không phải số nguyên

C

C

C

 A = 1 - (3/4) + (3/4)² - (3/4)³ + ... - (3/4)^2009 + (3/4)^2010 

A.(3/4) = (3/4) - (3/4)² + (3/4)³ - (3/4)^4 +... - (3/4)^2010 + (3/4)^2011 

cộng 2 đẳng thức trên lại vế theo vế: 
A + A.(3/4) = 1 + (3/4)^2011 => 7A/4 = 1 + (3/4)^2011 

=> 7A = 4 + 4.(3/4)^2011 không là số nguyên => A không nguyên 

vậy A ko phải là số nguyên

cám ơn bạn... Nguyễn Phương Hiền Thảo

Thank you very much

Nguyễn Phương Hiền Thảo 

Làm sao bạn biết 7A=4+4(3/4)^2011 ko phải số nguyên

C

C

c

\(B=3+3^2+3^3+3^4+...+3^{50}\)

\(\Rightarrow3B=3^2+3^3+3^4+3^5+...+3^{51}\)

\(\Rightarrow2B=3^{51}-3\)

\(\Rightarrow B=\frac{3^{51}-3}{2}\)

\(C=4+4^2+4^3+4^4+...+4^{2018}\)

\(\Rightarrow4C=4^2+4^3+4^4+4^5+...+4^{2019}\)

\(\Rightarrow3C=4^{2019}-4\)

\(\Rightarrow C=\frac{4^{2019}-4}{3}\)

\(B=3+3^2+3^3+...+3^{50}\)

\(\Rightarrow3B=3^2+3^3+3^4+....+3^{51}\)

\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{51}\right)-\left(3+3^2+...+3^{50}\right)\)

\(\Rightarrow2B=3^{51}-3\)

\(\Rightarrow B=\frac{3^{51}-3}{2}\)

\(C=4+4^2+4^3+...+4^{2018}\)

\(\Rightarrow4C=4^2+4^3+4^4+....+4^{2019}\)

\(\Rightarrow4C-C=\left(4^2+4^3+4^4+...+4^{2019}\right)-\left(4+4^2+4^3+...+4^{2018}\right)\)

\(\Rightarrow3C=4^{2019}-4\)

\(\Rightarrow C=\frac{4^{2019}-4}{3}\)

         \(B=3+3^2+...+3^{50}.\)

\(\Leftrightarrow3B=3^2+3^3+....+3^{51}\)

\(\Leftrightarrow3B-B=3^2+3^3....+3^{51}-\left(3+3^2+3^3+...+3^{50}\right)\)

\(\Leftrightarrow2B=3^{51}-3\)

\(\Leftrightarrow B=\frac{3^{51}-3}{2}\)

       \(C=4+4^2+....+4^{2018}\)

\(\Leftrightarrow4C=4^2+4^3+....+4^{2019}\)

\(\Leftrightarrow4C-C=4^2+4^3+...+4^{2019}-\left(4+4^2+....+4^{2018}\right)\)

\(\Leftrightarrow3C=4^{2019}-4\)

\(\Leftrightarrow C=\frac{4^{2019}-4}{3}\)

xét B=-3/4+(3/4)^2-.......-(3/4)^n với n lẻ,n>=1  

=>-3/4.B=(3/4)^2-(3/4)^3+.........+(3/4)...  

trừ theo vế suy ra 7/4.B=-3/4-(3/4)^(n+1)  

=>7B=-3-(3/4)^n  

=>A=1+B=1-(3+(3/4)^n)/7

 do <0(3/4)^n <1  

suy ra 0< 3+(3/4)^n <7  

suy ra (3+(3/4)^n)/7 ko là số nguyên  

suy ra A ko nguyên

****

vì A=0 nên A không phải số nguyên

\(B=-3\left(\dfrac{1}{4}-\dfrac{1}{4^2}+\dfrac{1}{4^3}-\dfrac{1}{4^4}+...-\dfrac{1}{4^{100}}\right)\)

Đặt \(C=\dfrac{1}{4}-\dfrac{1}{4^2}+...-\dfrac{1}{4^{100}}\)

\(\Leftrightarrow C\cdot\dfrac{1}{4}=\dfrac{1}{4^2}-\dfrac{1}{4^3}+...-\dfrac{1}{4^{101}}\)

\(\Leftrightarrow C\cdot\dfrac{-3}{4}=\dfrac{-1}{4^{101}}-\dfrac{1}{4}=\dfrac{-1-4^{100}}{4^{101}}\)

\(\Leftrightarrow C=\dfrac{-4^{100}-1}{4^{101}}\cdot\dfrac{-4}{3}=\dfrac{4^{100}+1}{3\cdot4^{100}}\)

\(\Leftrightarrow B=\dfrac{-4^{100}-1}{4^{100}}\)