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Bạn xem lại đề. Thay $n=1$ thì biểu thức không chia hết cho 7 nhé.
\(=3^3.3^n+3.3^n+2^3.2^n+2^2.2^n=\)
\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)=30.3^n+12.2^n=\)
\(=6\left(5.3^n+2.2^n\right)⋮6\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\left(9+3\right)+2^{n+2}\left(8+4\right)\)
\(=12.3^{n+1}+12.2^{n+2}=12.\left(3^{n+1}+2^{n+2}\right)\)
mà 12⋮6
\(\Rightarrow12.\left(3^{n+1}+2^{n+2}\right)⋮6\Rightarrow dpcm\)
a) \(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{14}\cdot2\cdot33⋮66\)
b) \(3^{m+2}-2^{n+4}+3^m+2^n\)
\(=3^m\cdot9+3-2^n\left(2^4-1\right)\)
\(=3^m\cdot10-2^{n-1}\cdot30\)
\(=30\left(3^{m-1}-2^{n-1}\right)⋮30\)
a) \(A=16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33=2^{14}\cdot66⋮66\)
b) Sửa đề
\(B=3^{n+2}-2^{n+4}+3^n+2^n=3^n\left(3^2+1\right)-2^n\left(2^4-1\right)=3^n\cdot10-2^n\cdot15\\ =3^{n-1}\cdot30-2^{n-1}\cdot30=30\left(3^{n-1}-2^{n-1}\right)⋮30\)
(với mọi n nguyên dương)
Ta có :
\(3^{n+2}-2^{n+2}+3^n-2^n\) =\(3^n.3^2-2^n.2^2+3^n-2^n\)
=\(3^n.9-2^n.4+3^n-2^n\) =\(3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
=\(3^n.10-2^n.5=3^n.10-2^{n-1}.2.5\) = \(3^n.10-2^{n-1}.10\)
=\(10.\left(3^n-2^{n-1}\right)⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\) (ĐPCM)
Sửa : 3n+2-2n+2+3n-2n
= 3n.9 - 2n.4+3n-2n
= 3n.10 - 2n.5
= 3n.10 - 2n.1/2.10
= 10 . (3n-2n.1/2) chia hết cho 10
Ta có : \(2n^3-6n^2-2n+n^2-3n-1-2n^3+1\)
=> \(-5n^2-5n=-5\left(n^2+n\right)\)Như vậy luôn chia hết cho 5 với mọi n
3n+2-2n+2 +3n-2n
=(3n+2+3n)+(-2n+2 -2n)
=3n.(32+1)-2n.(22+1)
=3n.10-2n.5
=3n.10-2n-1.10
=10.(3n-2n-1)chia hết cho 10
Vậy 3n+2-2n+2 +3n-2n chia hết cho 10
Ta có: \(3^{n+2}-2^{2n+4}+3^n+2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+4}-2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^4-1\right)\)
\(=3^n.10-2^n.15\)
\(=3^{n-1}.3.10-2^{n-1}.2.15\)
\(=3^{n-1}.30-2^{n-1}.30\)
\(=30\left(3^{n-1}-2^{n-1}\right)\)
Vì \(30⋮30\Rightarrow30\left(3^{n-1}-2^{n-1}\right)⋮30\)
\(\Rightarrow3^{n+2}-2^{n+4}+3^n+2^n⋮30\)
\(\Rightarrowđpcm\)
\(3^{n+2}-2^{n+4}+3^n+2^n\)
\(=3^n.3^2-2^n.2^4+3^n+2^n\)
\(=3^n\left(3^2+1\right)-2^n\left(2^4-1\right)\)
\(=3^n.10-2^n.15\)
mà 3n.10 \(⋮\)3.10=30
2n.15\(⋮\)2.15=30
\(\Rightarrow3^n.10-2^n.15⋮30\)
hay 3n+2-2n+4+3n+2n\(⋮\)30