Help with this proof (Index Sifting)

Question

Let $(x_j)^\infty_{j=1}$ be a sequence in $\mathbb{Z}$ and let $a, b, r \in \mathbb{Z}$ such that $a\le b$. Then

$\sum\limits_{j=a}^b x_j = \sum\limits_{j=a+r}^{b+r} x_{j-r}$

I assume that I would use induction to solve this, maybe on r?

Answer 1

Perhaps rewriting it as $\displaystyle\sum_{i=a}^b x_i = \sum_{j=a+r}^{b+r} x_{j-r}$ helps. Start with the summation on the left and make the substitution $j=i+r$.

Answer 2

Simply change the index so let $j=k-r\iff k=j+r$ so since $j\in\{a,\ldots,b\}$ then $k\in\{a+r,\ldots,b+r\}$ hence $$\sum_{j=a}^b x_j=\sum_{k=a+r}^{b+r}x_{k-r}$$ and remember that the variables are dummy so replace in the last sum $k$ by $j$ to find your result.