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
Yes, you can prove the claim by induction. For that, you must first analyse what the claim even is. In your case, you only have one variable, $n$ ($i$ is bound inside the sum), so clearly, induction will run on $n$, and the claim $P(n)$ will be
$$\sum_{i=1}^n a_i = \sum_{i=2}^{n+1} a_{i-1}$$
Also, for this proof, you will need to remember the definition of what $$\sum_{i=1}^n a_i$$ is. Remember that this expression is defined recursively, i.e.
The induction has two parts.
Part 1:
You need to prove $P(1)$. This part should be really easy. All you have to do is correctly write out what $P(1)$ means, and use the definitions of sums. In particular, you will use property 1 from above in this step.
Part 2:
Here, you assume that $P(n)$ is true, and you prove the statement $P(n+1)$. Again, correctly writing out $P(n)$ and $P(n+1)$ will get you 99% of the way there. In particular, in this step, property 2 from above will be useful.
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.
