Answer:
![a_n = 2^n -1](https://img.qammunity.org/2020/formulas/mathematics/college/v222c570j2ravsz2yixhf0p7pv98lvba5x.png)
Explanation:
We are given the following information:
![a_n = 2a_(n-1) + 1, a_0 = 0](https://img.qammunity.org/2020/formulas/mathematics/college/pk7vpn3cwj4fpl40bwm5ecgjuhbh63etak.png)
We will now evaluate values for n =1, 2, 3, 4, 5 and so on.
By forward substitution method:
![a_1 = 2a_0 + 1 = 0 + 1 = 1\\a_2 = 2a_1 + 1 = 2(1) + 1 =3\\a_3 = 2a_2 + 1 = 2(3) + 1 = 7\\a_4 = 2a_3 + 1 = 2(7) + 1 = 15\\a_5 = 2a_4 + 1 = 2(15) + 1 =31](https://img.qammunity.org/2020/formulas/mathematics/college/g8409iup389hvfuuej4eufaua2z14nrzmw.png)
If we continue in this manner, we can see a general trend and we can say that
![a_n = 2^n -1](https://img.qammunity.org/2020/formulas/mathematics/college/v222c570j2ravsz2yixhf0p7pv98lvba5x.png)
This is the solution for given recurrence relation.