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Đặt \(A=3+3^2+3^3+...+3^{15}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{13}+3^{14}+3^{15}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{13}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{13}.13\)
Vì \(13⋮13\)nên \(3.13+3^4.13+...+3^{13}.13⋮13\)
hay \(A⋮13\)
Vậy \(A⋮13.\)
A=3+3^2+3^3+......+3^13+3^14+3^15
=(3+3^2+3^3)+......+(3^13+3^14+3^15)
=3(1+3+3^2)+.......+3^13(1+3+3^2)
=(3+....+3^13)+(1+3+3^2)
=13(3+.....+3^13) chia hết cho 13
A=(1+3+3^2)+(3^3+3^4+3^5)+...........+(3^2013+3^2014+3^2015)
A=13+3^3.13+.......+3^2013.13
A=13.(1+3^3+....+3^2013)
vì 13chia hết cho 13
=>13.(1+3^3+......+3^2013) chia hết cho 13
hay A chia hết cho 13
\(A=1+3+3^2+3^3+3^4+...+3^{2015}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2013}+3^{2014}+3^{2015}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{2013}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\left(1+3^3+...+3^{2013}\right)\)
\(=13\left(1+3^3+...+3^{2013}\right)\)
Vì \(13\) chia hết cho 13 nên \(13\left(1+3^3+...+3^{2013}\right)\)chia hết cho 13
Vậy A chia hết cho 13
Ta có :
\(M=3^3+3^4+.....+3^{15}+3^{16}\)
\(\Rightarrow M=3^3\left(1+3\right)+......+3^{15}\left(1+3\right)\)
\(\Rightarrow M=3^3.4+......+3^{15}.4\)
=> M chia hết cho 4 .
\(M=\left(3^3+3^5\right)+....+\left(3^{14}+3^{16}\right)\)
\(\Rightarrow M=3^3\left(1+9\right)+.....+3^{14}\left(1+9\right)\)
\(\Rightarrow M=3^3.10+.....+3^{14}.10\)
=> M chia hết cho 10
A=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^97+3^98+3^99)
A=3.(1+3+3^2)+3^4.(1+3+3^2)+...+3^97.(1+3+3^2)
A=3.13+3^4.13+...+3^97.13
A=13.(3+3^4+...+3^97) chia hết cho 13
\(A=3+3^2+3^3+....+3^{99}\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\)
\(A=3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{97}.\left(1+3+3^2\right)\)
\(A=3.13+3^4.13+....+3^{97}.13\)
\(A=13.\left(3+3^4+....+3^{97}\right)\)
\(\Leftrightarrow A⋮13\)
Vậy: \(A⋮13\)
Nhớ k cho mình nhé! Thank you!!!
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
Vì 13 là lẻ \(\Rightarrow\) 13, 132, 133, 134, 135, 136 là lẻ.
Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 132 + 133 + 134 + 135 + 136 là chẵn. \(\Rightarrow\) 13 + 132 + 133 + 134 + 135 + 136 \(⋮\) 2
\(\Rightarrow\) ĐPCM
A=3+3^2+3^3+....+3^13+3^14+3^15
=(3+3^2+3^3)+...+(3^13+3^14+3^15)
=3(1+3+3^2)+...+3^13(1+3+3^2)
=(1+3+3^2)(3+...+3^13)
=13(3+...+3^13) chia hết cho 13