Prove $\sum_{i=1}^n a_i$ = $\sum_{i=2}^{n+1} a_{i-1}$ Given $\sum_{i=1}^n a_i$ = $\sum_{i=2}^{n+1} a_{i-1}$
How would you show this true for all n ∈ N and $a_1, a_2, . . . , a_n$ ∈ R?
I know it is obviously true because i would just use a substitution like i=j-1 then summing j-1 from 2 to n+1 gives the same result but not really sure how to show this. It seems like an induction problem to me but im not sure, how is this done? Thanks

Prove that 
$F(n):= \sum_{i=1}^{n}a_i-\sum_{i=2}^{n+1}a_{i-1}= 0$ , for $n \in \mathbb{Z^+}$ by induction.
1) $n=1$√.
2) Hypothesis $F(n)=0$.
3) Step for $n+1$.
$F(n+1)=$
$\sum_{i=1}^{n+1}a_i - \sum_{i=2}^{n+2}a_{i-1}=$
$\sum_{i=1}^{n}a_i +a_{n+1}$
$- \sum_{i=2}^{n+1}a_{i-1}- a_{n+1}=$
$F(n)+(a_{n+1}-a_{n+1})=0$,
since $F(n)=0$ by hypothesis, and the second summand is zero.
