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\(1+4+4^2+4^3+.....+4^{2018}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+....+\left(4^{2016}+4^{2017}+4^{2018}\right)\)
\(=21+\left[4^3\left(1+4+4^2\right)\right]+....+\left[4^{2016}\left(1+4+4^2\right)\right]\)
\(=21+4^3\cdot21+....+4^{2016}\cdot21\)
\(=21\left(1+4^3+....+4^{2016}\right)\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{120}\)
\(+\)Ta thấy \(B\)có số hạng là: \(\left(120-1\right):1+1=120\)(số hạng)
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Ta có: \(2^1+2^2+...+2^{2010}\)
\(=2\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=3\cdot\left(2+...+2^{2009}\right)⋮3\)
Ta có: \(2^1+2^2+...+2^{2010}\)
\(=2\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{2008}\right)⋮7\)
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`#3107.101107`
\(B=4+4^2+4^3+...+4^{89}+4^{90}\)
\(=\left(4+4^2+4^3\right)+...+\left(4^{88}+4^{89}+4^{90}\right)\)
\(=4\left(1+4+4^2\right)+...+4^{88}\left(1+4+4^2\right)\)
\(=\left(1+4+4^2\right)\left(4+...+4^{88}\right)\)
\(=21\left(4+4^{88}\right)\)
Vì \(21\left(4+4^{88}\right)\) `\vdots 21`
`\Rightarrow B \vdots 21`
Vậy, `B \vdots 21.`
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4+42+43+...+426
=(4+42)+...+(425+426)
=4.(1+4)+...+425.(1+4)
=4.5+...+425.5
=5.(4+...+425) CHIA HẾT CHO 20 VÀ K CHIA HẾT CHO 21
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Answer:
\(A=4+4^2+4^3+4^4+...+4^{99}\)
\(=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{96}+4^{97}\right)+\left(4^{98}+4^{99}\right)\)
\(=1\left(4+4^2\right)+4^2\left(4+4^2\right)+...+4^{95}\left(4+4^2\right)+4^{97}\left(4+4^2\right)\)
\(=1.20+4^2.20+...+4^{95}.20+4^{97}.20\)
\(=20.\left(1+4^2+...+4^{95}+4^{97}\right)\)
\(=5.4\left(1+4^2+...+4^{95}+4^{97}\right)⋮5\)
\(\Rightarrow A⋮5\)
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4 : 3 = tứ : tam = tám : tư = 8 : 4 = 2 (đpcm)
Ai k mk mk k lại !!
4 : 3 là tứ chia tam là tám chia tư 8 : 4 = 2