So I'm doing the binomial theorem in class at the moment, and I was wondering; when proving the binomial theorem by induction, the following steps occur:
$$(a+b)^{k+1}=\sum_{r=0}^{k} {k\choose r} a^{k-r+1}b^{r}+\sum_{r=0}^{k} {k\choose r}a^{k-r}b^{r+1}$$ let $s=r+1$, hence $r=s-1$ and $$=\sum_{r=0}^{k} {k\choose r}a^{k-r+1}b^{r}+\sum_{s=1}^{k+1} {k\choose s-1}a^{k-s+1}b^{s}$$ $$={k \choose 0}a^{k+1}b^0+\sum_{r=1}^k{k\choose r}a^{k-r+1}b^r+\sum_{s=1}^k {k\choose s-1}a^{k-s+1}b^s + {k\choose k}a^0b^{k+1}$$ $$=\sum_{r=1}^{k} {k\choose r}a^{k+1-r}b^k +a^{k+1} + b^{k+1}$$ $$=\sum_{r=0}^{k+1} {k+1\choose r}a^{k+1-r}b^{r}$$
and I definitely need some clarification on why the last two steps are possible. I understand how the combination would become $\displaystyle {k+1\choose r}$, but I'm struggling to understand how $\displaystyle\sum_{r=1}^{k} = \sum_{s=1}^{k+1}$ since $s\neq r$, unless they're made to be equal? But to me that wouldn't make any sense either since this would have complicated implications for the prior parts of the proof, would it not?
Any help is appreciated. Thank you.
Do the following simple trick: when $r$ is summed between $0$ and $k$, it means that $$ 0\leq r\leq k. $$ Now put $r=s-1$. Substitute above and obtain $$ 0\leq s-1\leq k, $$ or equivalently, isolating $s$, $$ 1\leq s\leq k+1. $$