Prove $\sum_{i=1}^n a_i$ = $\sum_{i=2}^{n+1} a_{i-1}$

Question

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

Answer 1

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.

1. $\sum_{i=1}^1 a_i = a_1$
2. $\sum_{i=1}^n a_i = \sum_{i=1}^{n-1} a_i + a_n$

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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.

Answer 2

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.