\(3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}\)|
\(=3^n\cdot3^3+3^n\cdot3+2^n\cdot2^2+2^n\cdot2\)
\(=3^n\left(3^3+3\right)+2^n\left(2^2+2\right)\)
\(=3^n\cdot30+2^n\cdot6\)
Vì 30 chia hết cho 6 nên 3n . 30 cũng chia hết cho 6.
Vì 6 chia hết cho 6 nên 2n .6 cũng chia hết cho 6.
Vậy .....
=))
Ta có:
\(A=3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}\)
\(=3^{n+1}\cdot3^2+3^{n+1}+2^{n+1}\cdot2^1+2^{n+1}\)
\(=3^{n+1}\cdot\left(3^2+1\right)+2^{n+1}\cdot\left(2^1+1\right)\)
\(=3^{n+1}\cdot10+2^{n+1}\cdot3\)
\(=3^n\cdot3\cdot2\cdot5+2^n\cdot2\cdot3\)
\(=3^n\cdot6\cdot5+2^n\cdot6\)
\(=6\cdot\left(3^n\cdot5\cdot2^n\right)\Rightarrow⋮6\left(đpcm\right)\)
3n+3+3n+1+2n+2+2n+1
=3n.33+3n.3+2n.22+2n.2
=3n.(33+3)+2n.(22+2)
=3n.30+2n. 6
3n.30 luôn chia hết cho 6 và 2n.6 luôn chia hết cho 6
=> 3n+3+3n+1+2n+2+2n+1 chia hết cho 6
\(3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}\)
\(=3^n.3^3+3^n.3+2^n.2^2+2^n.2\)
\(=3^n.\left(3^3+3\right)+2^n.\left(2^2+2\right)\)
\(=3^n.\left(27 +3\right)+2^n.\left(4+2\right)\)
\(=3^n.30+2^n.6\)
Vì : \(\hept{\begin{cases}3^n.30⋮6\\2^n.6⋮6\end{cases}}\)( Vì trong trong 2 tích có thừa số chia hết cho 6 )
\(\Rightarrow\)\(3^{n+3}+3^{n+1}+2^{n+2}+2^{n+1}⋮6\left(đpcm\right)\)
Rất vui vì giúp đc bn !!!