I am learning how to prove claims about finite sequences right now. Can you help me prove or disprove the following claim?
For every n where n greater than or equal to 1, for every
finite sequence a(0), a(1),...a(n) of n+1 elements, it holds that
$$\sum_{j=1}^n (a(j)-a(j-1)) = a(n)-a(0)$$
I know I need to prove this by induction but not sure what my hypothesis and inductive step should look like. Any ideas on how to prove this?
Base case $(n=1)$: \begin{align*} \sum_{j=1}^1 (a(j)-a(j-1)) &= a(1)-a(0) \\ a(1)-a(1-1) &= \\ a(1)-a(0) &\overset{\checkmark}{=} \end{align*}
Induction $(n>1)$:
Hypothesis: For all $m<n$, we have that $\sum_{j=1}^{m} (a(j)-a(j-1)) = a(m)-a(0)$. Then \begin{align*} \sum_{j=1}^{n} (a(j)-a(j-1)) &= (a(n)-a(n-1)) + \sum_{j=1}^{n-1} (a(j)-a(j-1)) \\ &= (a(n)-a(n-1)) + (a(n-1)-a(0)) \\ &= a(n) - a(n-1) + a(n-1) - a(0) \\ &= a(n) - a(0) \end{align*} which was to be shown.