Question

 help mee

 

Answer 1

\(a.A=2+2^2+2^3+...+2^{2017}\\ 2A=2^2+2^3+2^4+...+2^{2018}\\ 2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\\ A=2^{2018}-2\\ b.B=1+3^2+3^3+...+3^{2018}\\ 3B=3+3^3+3^4+...+3^{2019}\\ 3B-B=\left(3+3^3+3^4+..+3^{2019}\right)-\left(1+3^2+3^3+...+3^{2017}\right)\\ 2B=\left(3+3^{2019}\right)-\left(1+3^2\right)\\ 2B=3^{2019}-7\\ B=\dfrac{3^{2019}-7}{2}\)

Answer 2

`a)A = 2 + 2^2 + 2^3 + ...  + 2^2017`

`=> 2A = 2(2 + 2^2 + 2^3 + ... + 2^2017)`

`=>2A = 2^2 + 2^3 + 2^4 + .... + 2^2018`

`-`

` A = 2 + 2^2 + 2^3 + ... + 2^2017`

`=> A = 2^2018 - 2`

Vậy `A = 2^2018 - 2`

`b)B = 1 + 3^2 + 3^4 + ... + 3^2018`

`=> 3B = 3(1 + 3^2 + 3^4 + ... + 3^2018)`

`=> 3B = 3 + 3^3 + 3^5 + ... + 3^2019`

`-`

`B = 1 + 3^2 + 3^4 + ... + 3^2018`

`=> 2B = (3 + 3^2019) - (1 + 3^2)`

`=> 2B = 3+ 3^2019 - 1 -3^2`

`=> 2B= 3^2019 - 7`

`=> B = (3^2019 - 7)/2`

Vậy B = (3^2019 - 7)/2`