Answer
$$\sum^{\infty}_{n=0}a_{n}(x-1)^{n+1}$$
Work Step by Step
1. Add 1 to each n in the second expression:
$$\sum^{\infty}_{n+1=1}a_{n+1-1}(x-1)^{n+1} =\sum^{\infty}_{n=1-1}a_{n}(x-1)^{n+1} $$ $$\sum^{\infty}_{n=0}a_{n}(x-1)^{n+1}$$
- We get exactly the first expression, therefore, they are equivalent.