`u_{n+2} = (5u_{n+1} - 2u_{n})/3`

`to 3u_{n+2} = 5u_{n+1} - 2u_{n}`

 `to 3u_{n+2} - 3u_{n+1} = 2u_{n+1} - 2u_{n}`

`to 3(u_{n+2} - u_{n+1}) = 2(u_{n+1} - u_{n})`

Đặt `v_{n} = u_{n+1} - u_{n}`

`to {(v_{1} = u_{2020} - u_{2019} = 1),( q = 2/3):}`

`to v_{n} = v_{1} . q^(n - 1) = (2/3)^(n-1)`

`u_{n} - u_{n-1} = v_{n-1} = (2/3)^(n-2)`

`u_{n-1} - u_{n-2} = v_{n-2} = (2/3)^(n-3)`

.....

`u_{2}-u_{1} = v_{1} = 1`

`to u_{n} - u_{n - 1} + u_{n - 1} - u_{n-2} + ... + u_{2} - u_{1} = (2/3)^(n-2) + (2/3)^(n-3) + .... + 1`

`to u_{n} - u_{1} = ( 1 - (2/3)^(n-1))/(1 - 2/3) = 3 - 3 . (2/3)^(n-1) = 3 . ( 1 - (2/3)^(n-1))`

`to u_{n} = 3 . (1 - (2/3)^(n-1) ) + u_{1} = 3 . (1 - (2/3)^(n-1)) + 2019`

`to a = 3 ; b = 2 ; c = 3 ; d = 2019`

`to a + 3b - 3c + d = 3 + 3 . 2 - 3 . 3 + 2019 = 2019`