Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)

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

Ta có: \(3^{n+2}+3^n=3^n\left(3^2+1\right)=10.3^n⋮10\)

\(2^{n+2}+2^n=2^n\left(4+1\right)=5.2^n=10.2^{n-1}⋮10\)

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

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

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

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

A = 3\(^{n}\).(9 + 1) - 2\(^{n}\).(2\(^3\) + 2)

A = 3\(^{n}\).10 - 2\(^{n}\).(8 + 2)

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

A ⋮ 10(đpcm)

\(3^{n+3}+2^{n+3}-3^{n+2}+2^{n+2}=27.3^n-9.3^n+8.2^n+4.2^n\)

\(=3^n\left(27-9\right)+2^n\left(8+4\right)\)

\(=6.3^{n+1}+6.2^{n+1}\)

\(=6\left(3^{n+1}+2^{n+1}\right)⋮6\left(đpcm\right)\)

= (3ⁿ⁺²  + 3ⁿ ) - (2ⁿ⁺² +  2ⁿ) = 3ⁿ. (3² + 1) - 2ⁿ .(2² + 1) = 3ⁿ.10 - 2ⁿ .5 = 3ⁿ.10 - 2^n-1.2 .5 =  10. (3ⁿ - 2^n-1)  chia hết cho 10 

=>  3ⁿ⁺²  + 3ⁿ  - 2ⁿ⁺² +  2ⁿ chia hết cho 10 với mọi n

bài làm

= (3ⁿ⁺²  + 3ⁿ ) - (2ⁿ⁺² +  2ⁿ)

= 3ⁿ. (3² + 1) - 2ⁿ .(2² + 1)

= 3ⁿ.10 - 2ⁿ .5 

= 3ⁿ.10 - 2^n-1.2 .5

=  10. (3ⁿ - 2^n-1)  chia hết cho 10 

Vậy ..................

hok tốt

\(3^{n+2}+3^n-\left(2^{n+2}+2^n\right)\)\(=9.3^n+3^n-\left(8.2^{n-1}+2^{n-1}\right)=10.3^n-10.2^{n-1}\)

chia hết cho 10

\(=9.3^n+3^n-8.2^{n-1}-2.2^{n-1}=10.3^n-10.2^{n-1}\)

chia hết cho 10