cho b =\(^{3^1}\)+\(^{3^2}\)+\(3^3\).............+\(^{3^{300}}\) chứng minh b chia hết cho 2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=4^1+4^2+...+4^{300}\)
\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{299}\left(1+4\right)\)
\(=4.5+4^3.5+...+4^{299}.5=5\left(4+4^3+...+4^{299}\right)⋮5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=4+4^2+4^3+...+4^{300}\)
\(B=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{299}+4^{300}\right)\)
\(B=5.4+5.4^3+...+5.4^{299}\)
\(B=5\left(4+4^3+4^5+...+4^{299}\right)\)
\(\Rightarrow B⋮5\)
\(\sqrt{\sqrt[]{}\sqrt[]{}\begin{matrix}&\\&\\&\end{matrix}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{299}+3^{300}\right)\\ S=\left(1+3\right)\left(1+3^2+...+3^{299}\right)\\ S=4\left(1+3^2+...+3^{299}\right)⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B= 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 +...+ 298 - 299 - 300 + 301 + 302
= 1 + ( 2 - 3 - 4 + 5) + ( 6 - 7 - 8 + 9) + ( 10 - 11 - 12 + 13) +...+ (298 - 299 - 300 + 301 ) + 302
= 1 + 0 + 0 +...+ 0 + 302
= 1 + 302 = 303 chia hết cho 3
=> B chia hết cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
1: \(2A=2+2^2+...+2^{2011}\)
=>\(A=2^{2011}-1>B\)
2: \(A=\left(2010-1\right)\left(2010+1\right)=2010^2-1< B\)
3: \(A=1000^{10}\)
\(B=2^{100}=1024^{10}\)
mà 1000<1024
nên A<B
5: \(A=3^{450}=27^{150}\)
\(B=5^{300}=25^{150}\)
mà 27>25
nên A>B
b=31+32+...+3300
b=(3+32)+(33+34)+...+(3299+3300)
b=3(1+3)+33(1+3)+...+3299(1+3)
b=4(3+33+...+3299)
b=2.2(3+33+...+3299)
\(\Rightarrow\)b\(⋮\)2
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