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\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)
\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(b,B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{101}-1}{2}\)
\(c,C=4+4^2+4^3+...+4^n\)
\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)
\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)
\(3C=4^{n+1}-4\)
\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)
\(d,D=1+5+5^2+...+5^{2000}\)
\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(\Rightarrow D=\frac{5^{2001}-1}{4}\)
b)
B=1+3+3^2+3^3+..+3^100
=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101
=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)
=> 2B = 3^101 - 1
=> B =( 3^101 - 1) / 2
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
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
\(a,2010:\left(-5\right)+400-1\\ =-402+400-1\\ =-3\\ b,\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)\\ =\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c,\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\\ =\dfrac{1}{12}.\left(\dfrac{1}{20}\right)^2\\ =\dfrac{1}{12}.\dfrac{1}{400}\\ =\dfrac{1}{4800}\)
a) \(2010:\left(-5\right)+400-1=-400+400-1=-1\)
b) \(\dfrac{2}{3}+\dfrac{3}{4}\cdot\dfrac{-4}{9}=\dfrac{2}{3}+\dfrac{-1}{3}=\dfrac{1}{3}\)
c) \(\left(1-\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{1}{12}\cdot\dfrac{1}{400}=\dfrac{1}{4800}\)
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
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