a) 8 . 2ⁿ + 2ⁿ⁺¹ = 2ⁿ . ( 8 + 2 ) = 2ⁿ . 10 = ....0 

b) có vấn đề

c) 4ⁿ⁺³ + 4ⁿ⁺² - 4ⁿ⁺¹ - 4ⁿ = 4ⁿ . ( 4³ + 4² - 4 - 1 ) = 4ⁿ . 75 = 4^n-1 . 4 . 75 = 300 . 4^n-1 \(⋮\)300

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

Đây nè bạn

=>(3^n+2)+(3^n)-(2^n+2)-(2^n)=3^n((3^2)+1)-2^n((2^2)+1)=(3^n)*10-(2^n)*5=(3^n)*10-(2^n-1)*5*2=(3^n)*10-(2^n-1)*10=10*((3^n)-(2^n-1) chia hết cho 10

=>(3^n+2)-(2^n+2)+(3^n)-(2^n)chia hết cho 10

\(5^{n+2}+3^{n+2}-3^n-5^n=5^n\left(5^2-1\right)+3^n\left(3^2-1\right)=5^n.24+3^n.8\)

Ta có \(5^n.24⋮24\) và \(3^n.8⋮3.8=24\)

Vậy ta đc đpcm

Ta có  và 

\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10.\left(3^n-2^{n-1}\right)⋮10\) ∀n∈N

Vậy ...

Tham khảo

=\(3^n\).\(3^2\)-\(2^n\).\(2^2\)+\(3^n\)-\(2^n\)

=\(^{3^n}\).9 - \(2^n\).4 +\(^{3^n}\)- \(2^n\)

=10 .\(3^n\)-5.\(2^n\)

=10.\(3^n\)-5.2.\(2^{n-1}\)

=10 .(\(3^n\)-\(2^n\) )

=> chia hết cho 10

Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)\)

\(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)

\(=3^n\cdot10-2^n\cdot5\)

\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)

\(=3^n\cdot10-2^{n-1}\cdot10\)

\(=\left(3^n-2^{n-1}\right)\cdot10⋮10\left(dpcm\right)\)

Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n=3^{n+2}+3^n-\left(2^{n+2}+2^n\right)\)

Thấy: \(3^{n+2}+3^n=3^n.2^2+3^n=9.3^n+3^n=3^n.\left(9+1\right)=3^n.10\)

\(\Rightarrow3^{n+2}+3^n⋮10\)\(\left(1\right)\)

\(2^{n+2}+2^n=4.2^n+2^n==2^n\left(4+1\right)=2^n.5=2.2^{n-1}.5=10.2^{n-1}\)

\(\Rightarrow2^{n+2}+2^n⋮10\)\(\left(2\right)\)

Từ (1) và (2) \(\Rightarrow3^{n+2}+2^n-\left(2^{n+2}+2^n\right)⋮10\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\) (đpcm)

k!