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3^n+2-2^n+2+3^n-2^n
=3^n+2+3^n-(2^n+2+2^n)
=3^n(3^2+1)-2^n(2^2+1)
=3^n.10-2^n.5=3^n.10-2^n-1.10=10(3^n-2^n-1) chia hết cho 10(đpcm)
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Ta có:
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5\)
Vì \(\left\{{}\begin{matrix}3^n.10⋮10\\2^n.5⋮10\end{matrix}\right.\)
Nên \(3^{n+2}-2^{n+2}+3^n-2^n\) chia hết cho 10
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\(3^{n+2}-2 ^{n+2}+3^n-2^n=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1}.10=\left(2^n-2^{n-1}\right).10\) chia hết cho 10
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Đăt S = 3^(n+2)-2^(n+2)+3^n-2^n = 3^(n+2) + 3^n - [2^(n+2) + 2^n]
Ta có 3^(n+2) + 3^n = 9.3^n + 3^n = 10.3^n (chia hết cho 10)
Và 2^(n+2) + 2^n = 4.2^n + 2^n = 5.2^n (chia hết cho 10, vì chia hết cho 2 và 5)
Suy ra S chia hết cho 10.
\(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-2^{n+2}-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)=3^n.10-2^n.5=3^n.10-2^{n-1}.10=\left(3^n-2^{n-1}\right).10\)
luôn chia hết cho 10 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=4^{n+3}+4^{n+2}-4^{n+1}-4^n\)
\(A=4^{n-1}\left(4^4+4^3-4^2-4\right)\)
\(A=4^{n-1}.\left(300\right)\)
\(A=4^{n-1}.\left(300\right)⋮300\)
Vậy...
hok tốt!!!
\(3^{n+2}-2^{n+2}+3^n-2^n=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1}.10=\left(3^n-2^{n-1}\right).10\) chia hết cho 10