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