I had a question with regards to the final step in proving the binomial theorem using induction. After changing indices on the second term I arrive at the following:
$$\sum_{j = 0}^{k}\binom{k}{j}a^{k+1-j}b^{j} + \sum_{j = 1}^{k + 1}\binom{k}{j - 1}a^{k+1-j}b^{j}$$
It is the following step that I'm having an issue with. I know that $\binom{k+1}{j} = \binom{k}{j} + \binom{k}{j-1}$.
I'm having trouble justifying to myself the next steps. Now what I would like to do is group everything together under one summation and then factor out the shared terms:
$$ \sum_{j = 0}^{k+1}\bigg[\binom{k}{j}a^{k+1-j}b^{j} + \binom{k}{j - 1}a^{k+1-j}b^{j}\bigg] = \sum_{j = 0}^{k+1}\bigg(\binom{k}{j} + \binom{k}{j-1}\bigg)a^{k+1-j}b^{j}$$
This would then result in the desired final result:
$$ \sum_{j = 0}^{k+1}\binom{k+1}{j}a^{k+1-j}b^{j}$$
My issue is with the indices. As can be seen when $j = 0$ I will end up with a negative value in my second binomial coefficient, not a good thing. So what or how do I reconcile that one piece to be able to have the factorization work out the way I envision?
The easiest thing to do is define $\binom{k}{j} = 0$ if $j < 0$ or $j>k$. That way $\binom{k+1}{j} = \binom{k}{j} + \binom{k}{j-1}$ is still true when $j=0$.
Otherwise, you have to restrict the statement of Pascal's identity: $$ \binom{k+1}{j} = \begin{cases} 1 & j=0 \\ 1 & j=k+1 \\ \binom{k}{j} + \binom{k}{j-1} & 1 \leq j \leq k \end{cases} $$