A summation proof

Question

Prove USING INDUCTION ($\mathbb{N}$ starting $1$) that: $\sum_{n=a}^{b}f(n) = \sum_{n=0}^{b-a} f(b-n)$

I found this property of summations in the Wikipedia page, but I haven't been able to prove it. Here's my attempt:

Let $a,b\in \mathbb{Z}$ and $a<b$.

1. Base case: if $b=a+1$, \begin{align*} \sum_{n=0}^{b-a} f(b-n) &= \sum_{n=0}^{(a+1)-a} f(a+1-n)\\ &= \sum_{n=0}^{1} f(a+1-n)\\ &= f(a+1-0) + \sum_{n=1}^{1} f(a+1-n)\\ &= f(a+1) + f(a+1-1)\\ &= f(a+1) + f(a)\\ &= f(a) + f(a+1)\\ &= \sum_{n=a}^{a+1} f(n)\\ &= \sum_{n=a}^{b} f(n) \end{align*}
2. Assume that it is true for $b=a+k$, for some $k\in \mathbb{Z}$, that is to say, supposed that: \begin{align*} \sum_{n=a}^{a+k}f(n) &= \sum_{n=0}^{(a+k)-a} f(a+k-n)\\ &= \sum_{n=0}^{k} f(a+k-n) \end{align*}
3. For $b=a+k+1$ we get: \begin{align*} \sum_{n=0}^{b-a}f(b-n) &= \sum_{n=0}^{(a+k+1)-a} f((a+k+1)-n)\\ &= \sum_{n=0}^{k+1} f(a+k+1-n)\\ &= \text{What goes next?} \end{align*}

that's as far as I get.

Answer 1

Consider f(b - n) for all $n\in[0, b-a]$

When n = 0, f(b - n) = f(b)

When n = 1, f(b - n) = f(b - 1)

…

When n = b - a, f(b - n) = f(a)

Thus you get all the same terms as on the right, but in reverse order. Since addition is associative, the sums are equal. $\blacksquare$

Answer 2

Suppose $\sum_{n=a}^{b}f(n) = \sum_{n=0}^{b-a} f(b-n) $ is true for $b-a=m$. This means $b=a+m$ so $\sum_{n=a}^{a+m}f(n) = \sum_{n=0}^{m} f(a+m-n) $

Then, if $b-a=m+1$, $b=a+m+1$ so

$\begin{array}\\ \sum_{n=a}^{b}f(n) &=\sum_{n=a}^{a+m+1}f(n)\\ &=\sum_{n=a}^{a+n}f(n)+f(a+m+1)\\ &=\sum_{n=0}^m f(n+a)+f(a+m+1)\\ &=\sum_{n=0}^m f(a+m-n)+f(a+m+1)\\ &=\sum_{n=-1}^m f(a+m-n)\\ &=\sum_{n=0}^{m+1} f(a+m-(n-1))\\ &=\sum_{n=0}^{m+1} f(a+m+1-n)\\ &=\sum_{n=0}^{b-a} f(b-n)\\ \end{array} $