Chứng minh rằng: A =\(3^{n+2}-2^{n+2}+3^n-2^n\) chia hết cho 10 với n\(\in N\)
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1)
a)251-1
=(23)17-1\(⋮\)23-1=7
Vậy 251-1\(⋮\)7
b)270+370
=(22)35+(32)35\(⋮\)22+32=13
Vậy 270+370\(⋮\)13
c)1719+1917
=(BS18-1)19+(BS18+1)17
=BS18-1+BS18+1
=BS18\(⋮\)18
d)3663-1\(⋮\)35\(⋮\)7
Vậy 3663-1\(⋮\)7
3663-1
=3663+1-2
=BS37-2\(⋮̸\)37
Vậy 3663-1\(⋮̸\)37
e)24n-1
=(24)n-1\(⋮\)24-1=15
Vậy 24n-1\(⋮\)15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{n+2}-2^{n+2}+3^n-2^n=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5=3^n.10-2^{n-1}.10=\left(3^n-2^{n-1}\right).10\) chia hết cho 10(đpcm)
tick tớ nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=10.3^n-5.2^n=10.3^n-5.2.2^{n-1}=10\left(3^n-2^{n-1}\right)⋮10\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
3n+2 -2n+2 +3n -2n
=3n .32 -2n .22 +3n -22
=3n(9+)-2n(4-1)
Vì 3n .10 ⋮10
=> 3n .10- 2n .3⋮10
=>3n +2 -2n+2 +3n -2n ⋮10
sai
trước 2^n là dấu trừ => trong ngoặc đổi dấu thành 2^n(4+1)
=>2^n-1.10 chia hết cho 10
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(\Rightarrow\left(3^n\cdot3^2+3^n\right)-\left(2^n\cdot2^2+2^n\right)\)
\(\Rightarrow3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(\Rightarrow3^n\cdot10-2^n\cdot5\)
\(\Rightarrow3^n\cdot10-2^{n-1}\cdot\left(2\cdot5\right)\)
\(\Rightarrow10\left(3^n-2^n\right)\) chia hết cho 10
b) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(\Rightarrow3^n\cdot3^3+3^n\cdot3+2^n\cdot2^3+2^n\cdot2^2\)
\(\Rightarrow3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)
\(\Rightarrow3^n\cdot30+2^n\cdot12\)
\(\Rightarrow3^n\cdot6\cdot5+2^n\cdot2\cdot6\)
\(\Rightarrow6\left(3^n\cdot5+2^n\cdot2\right)\) chia hết cho 6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{n+1}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)
\(A=3^{n+2}-2^{n+2}+3^n-2^n\)
=> \(A=3^n.3^2-2^n.2^2+3^n-2^n\)
=> \(A=3^n\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
=> \(A=3^n.10-2^n.5\)
=> \(A=3^n.10-2^{n-1}.2.5\)
=> \(A=3^n.10-2^{n-1}.10\)
=> \(A=10\left(3^n-2^{n-1}\right)\)
Vì 10 chia hết cho 10 => \(10\left(3^n-2^{n-1}\right)\) chia hết cho 10 => A chia hết cho 10
A=3^n+2-2^n+2+3^n-2^n
A=3^n.3^2-2^n.2^2+3^n-2^n
A=3^n.10-2^n.5
A=10(3^n-2^n-1)
=> A chia hết cho 10.