Summation operation for precalculus

Studying Spivak's Calculus I came across a relation I find hard to grasp. In particular, I want to understand it without using proofs by induction. So please prove or explain the following relationship by not using induction.

$$\sum_{j=0}^{n}\binom{n}{j}a^{n-j}b^{j+1}=\sum_{j=1}^{n+1}\binom{n}{j-1}a^{n+1-j}b^{j}$$

Thanks in advance.

Answer

The identity you've given appears to be an index shift. Instead of beginning to sum at $i=0$, we wish to begin at $1$. In order to advance the summation index ahead by $1$, we have to take away $1$ from every instance of the index variable inside the summand.

$$\sum_{i=0}^{n}\binom{n}{i}a^{n-i}b^{i+1}$$

The index shift becomes clear if you let $j = i + 1$ and substitute.

$$= \sum_{j=0+1}^{n+1}\binom{n}{j-1}a^{n-(j-1)}b^{(j-1)+1}$$
$$= \sum_{j=1}^{n+1}\binom{n}{j-1}a^{n-j+1}b^{j}$$