Chứng minh rằng :
\(a.\)
\(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(b.\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 7 2016
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
6 tháng 7 2016
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
5 tháng 7 2019
\(B=\left(3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\right)\)
\(=3^{n+1}\left(3^2+1\right)-2^{n+1}\left(2^2+1\right)\)
\(=3^{n+1}.10-2^{n+1}.5\)
\(=3^{n+1}.10+2^n.2.5\)
\(=3^{n+1}.10+2^n.10\)
\(=10\left(3^{n+1}+2^n\right)\)\(⋮\)\(10\)\(\left(đpcm\right)\)
5 tháng 7 2019
\(Â=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+1}\)
\(=3^n\left(3^3+3\right)+2^{n+1}\left(2^2+1\right)\)
\(=3^n.30+2^{n+1}.\left(2^2+2\right).\frac{1}{2}\)
\(=3^n.30+2^{n+1}.6.\frac{1}{2}\)
Mà \(3^n.30⋮6;2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow3^n.30+2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow A⋮6\left(đpcm\right)\)
TT
2
1 tháng 9 2018
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\left(3^2+1\right)-2^{n+2}-2^n\)
\(=10.3^n-5.2^n\)
Do 2^n chia hết cho 2 suy ra 5.2^n chia hết cho 10 nên:
\(10.3^n-5.2^n⋮10\left(ĐCCM\right)\)
1 tháng 9 2018
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)
\(=30.3^n+12.2^n\)
\(=6\left(5.3^n+2^{n+1}\right)\)
MT
15 tháng 7 2015
a_)3n+2 - 2n+2 +3n - 2n
=(3n+2+3n)+(-2n+2-2n)
=(3n.32+3n.1)+(-2n.22-2n+1)
=3n.(9+1)-2n.(4+1)
=3n.10-2n.5
ta có 3n.10 chia hết cho 10 và 2n.5 chia hết cho 10( vì có thừa số 2 và 5)
=> 3n+2 - 2n+2 +3n - 2n chia hết cho 10.
\(.a.\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
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 : \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(.b.\) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
Ta có : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\) \(=3^n.\left(3^3+3\right)+2^n.\left(2^3+2^2\right)\)\(=3^n.30+2^n.12\)
\(=6\left(3^n.5+2^{n+1}\right)⋮6\) \(\left(dpcm\right)\)
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)\(VT=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=\left(3^{n+3}+3^{n+1}\right)+\left(2^{n+3}+2^{n+2}\right)\)
\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}\cdot10+2^{n+2}\cdot3\)
\(=3^n\cdot3\cdot2\cdot5+2^{n+1}\cdot2\cdot3\)
\(=3^n\cdot5\cdot6+2^{n+1}\cdot6\)
\(=6\cdot\left(3^n\cdot5+2^{n+1}\right)⋮6\)