\(A=3+3^2+3^3+...+3^{2015}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2015}+3^{2016}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2016}-3\right)\)

\(\Rightarrow2A=3^{2016}-3\)

\(\Rightarrow A=\dfrac{3^{2016}-3}{2}\)

Ta có: \(2A+3=3^n\)

\(\Rightarrow2\cdot\dfrac{3^{2016}-3}{2}+3=3^n\)

\(\Rightarrow3^{2016}-3+3=3^n\)

\(\Rightarrow3^{2016}=3^n\)

\(\Rightarrow n=2016\)

81^7 - 27^9 - 9^13
= (3^4)^7 - (3^3)^9 - (3^2)^13
= 3^28 - 3^27 - 3^26
= (3^26.3^2) - (3^26.3^1) - (3^26.1)
= 3^26.(9 - 3 - 1)
= 3^22.(3^4.5)
= 3^22.405 chia hết cho 405
=> 81^7 - 27^9-9^13 chia hết cho 405

Không chia hết đâu bạn ơi

Bạn xem lại đề. Thay $n=1$ thì biểu thức không chia hết cho 7 nhé.

\(\left(\dfrac{2}{5}-\dfrac{1}{3}\right)^2\)

=  \(\left(\dfrac{1}{15}\right)^2\)

=  \(\dfrac{1}{225}\)

\(=3^2.3^n-2^2.2^n+3^n-2^n=10.3^n-5.2^n=\)

\(=10.3^n-10.2^{n-1}=10.\left(3^n-2^n\right)⋮10\)

\(\left(3b^2\right)^2-b^3.\left(1-5b\right)\)

\(=9b^4-b^3+5b^4\)

\(=14b^4-b^3\)

4x\(^{1+n-2}\)y\(^{1+n+1}\)4xy\(^{1+n}\)+4xy

\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)

\(21⋮7\Rightarrow\left(5^3.21\right)⋮7\Rightarrow\left(5^5-5^4+5^3\right)⋮7\)