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Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)
`3^(n+2)-2^(n+2)+3^n-2^n`
`=3^n .3^2 -2^n .2^2+3^n-2^n`
`= 3^n .(3^2+1)-2^n .(2^2+1)`
`=3^n .10 - 2^n . 5`
`=3^n .10 - 2^(n-1) .2.5`
`=3^n .10 -2^(n-1) .10 vdots 10 `
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
Ta có:3n+3+3n+1+2n+3+2n+2=3n+1(32+1)+2n+2(2+1)=10.3n+1+2n+23=3.2.(5.3n+1+2n+1)chia hết cho 6
Vậy...
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\) ∀n∈N
Vậy ...
\(S=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=10.3^n-5.2^n=10.3^n-5.2.2^{n-1}=10\left(3^n-2^{n-1}\right)⋮10\) (đpcm)
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