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a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(S=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{27}+5^{28}\right)\)
\(S=1\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(S=\left(1+5^2+...+5^{27}\right).6⋮3\left(dpcm\right)\)
b) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{29}\)
\(\Rightarrow5S-S=\left(5+5^2+5^3+5^4+...+5^{29}\right)-\left(1+5+5^2+5^3+...+5^{28}\right)\)
\(\Rightarrow4S=5^{29}-1\)
\(\Rightarrow4S+1=5^{29}-1+1\)
\(\Rightarrow4S=5^{29}=5^n\)
\(\Rightarrow n=29\)
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow S=\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(\Rightarrow S=6+5^2.6+...+5^{27}.6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮3\)
\(\Rightarrow dpcm\)
b) Bạn xem lại đề
\(25A=\dfrac{5^{502}+25}{5^{502}+1}=1+\dfrac{24}{5^{502}+1}\)
\(25B=\dfrac{5^{602}+25}{5^{602}+1}=1+\dfrac{24}{5^{602}+1}\)
\(5^{502}+1< 5^{602}+1\)
=>\(\dfrac{24}{5^{502}+1}>\dfrac{24}{5^{602}+1}\)
=>25A>25B
=>A>B
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{98^2}\)
A=\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+\(\frac{1}{5.5}\)+...+\(\frac{1}{98.98}\)
A<\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{97.98}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{97}\)-\(\frac{1}{98}\)=\(\frac{1}{2}\)-\(\frac{1}{98}\)=\(\frac{24}{49}\)<1.
Vậy A<1