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Bổ sung điều kiện n ∈ N
\(3^{n+3}+2^{n+3}+3^{n+1}+2^{n+2}\)
\(=3^n\cdot3^3+2^n\cdot2^3+3^n\cdot3+2^n\cdot2^2\)
\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)
\(=3^n\cdot30+2^n\cdot12\)
Ta có : \(\hept{\begin{cases}3^n\cdot30⋮6\\2^n\cdot12⋮6\end{cases}}\Rightarrow3^n\cdot30+2^n\cdot12⋮6\)
=> \(3^{n+3}+2^{n+3}+3^{n+1}+2^{n+2}⋮6\)( đpcm )
\(3^{n+3}+2^{n+3}+3^{n+1}+2^{n+2}\)
\(=3^n.27+2^n.8+3^n.3+2^n.4\)
\(=3^n\left(27+3\right)+2^n\left(8+4\right)\)
\(=3^n.30+2^n.12\)
\(=6.\left(3^n.5+2^n.2\right)⋮6\)
\(B=\left(3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\right)\)
\(=3^{n+1}\left(3^2+1\right)-2^{n+1}\left(2^2+1\right)\)
\(=3^{n+1}.10-2^{n+1}.5\)
\(=3^{n+1}.10+2^n.2.5\)
\(=3^{n+1}.10+2^n.10\)
\(=10\left(3^{n+1}+2^n\right)\)\(⋮\)\(10\)\(\left(đpcm\right)\)
\(Â=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+1}\)
\(=3^n\left(3^3+3\right)+2^{n+1}\left(2^2+1\right)\)
\(=3^n.30+2^{n+1}.\left(2^2+2\right).\frac{1}{2}\)
\(=3^n.30+2^{n+1}.6.\frac{1}{2}\)
Mà \(3^n.30⋮6;2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow3^n.30+2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow A⋮6\left(đpcm\right)\)
Chứng minh rằng :
\(a.\)
\(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(b.\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(.a.\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Ta có : \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.\left(3^2+2\right)-2^n.\left(2^2+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\) \(\left(dpcm\right)\)
Vậy : \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(.b.\) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(=3^n.30+2^n.12\)
\(=6\left(3^n.5+2^{n+1}\right)⋮6\) \(\left(dpcm\right)\)
Vậy : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
a)\(VT=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\)
b)\(VT=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=\left(3^{n+3}+3^{n+1}\right)+\left(2^{n+3}+2^{n+2}\right)\)
\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}\cdot10+2^{n+2}\cdot3\)
\(=3^n\cdot3\cdot2\cdot5+2^{n+1}\cdot2\cdot3\)
\(=3^n\cdot5\cdot6+2^{n+1}\cdot6\)
\(=6\cdot\left(3^n\cdot5+2^{n+1}\right)⋮6\)
b: \(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)
c: \(=3^n\left(3^2+3\right)+2^n\left(2^3+2^2\right)\)
\(=3^n\cdot12+2^n\cdot12⋮6\)
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 cóA= 3n+3+2n+3+3n+1+2n+2=3n.27+2n.8+3n.3+2n.4=3n.(27+3)+2n.(8+4)=3n.30+2n.12
Vì 30 chia hết cho 6 ,12 chia hết cho 6 suy ra 3n.30 chia hết cho 6,2n.12 chia hết cho 6
suy ra 3n.30+2n.12 chia hết cho 6
suy ra A chia hết cho 6
Ta có : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}.10+2^{n+1}.3\)
\(=3^n.5.6+2^{n+1}.6⋮6\)
Lời giải:
Ta có: \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}=3^{n}.3^3+3^n.3+2^n.2^3+2^n.2^2\)
\(=3^n(3^3+3)+2^n(2^3+2^2)\)
\(=3^n.30+2^n. 12=6(3^n.5+2^n.2)\vdots 6\)
Ta có đpcm.
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^n.3^3+3^n.3+2^n.2^3+2^n.2^2\)
\(=3^n.\left(3^3+3\right)+2^n.\left(2^3+2^2\right)\)
\(=3^n.30+2^n.12\)
\(=3^n.5.6+2^n.2.6\)
\(=6.\left(3^n.5+2^n.2\right)\)
Vì \(6⋮6\)
\(\Rightarrow6.\left(3^n.5+2^n.2\right)⋮6\) \(\forall n\in N.\)
\(\Rightarrow3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\) \(\forall n\in N\left(đpcm\right).\)
Chúc bạn học tốt!
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...