I'm asked to use the fact that $\begin{pmatrix}n\\r\end{pmatrix}+\begin{pmatrix}n\\r-1\end{pmatrix}=\begin{pmatrix}n+1\\r\end{pmatrix}$ to show, by induction, that $$(a+b)^n=\begin{pmatrix}n\\0\end{pmatrix}a^n+\begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\cdots+\begin{pmatrix}n\\r\end{pmatrix}a^{n-r}b^r+\begin{pmatrix}n\\n\end{pmatrix}b^n,$$ but I'm not sure where to start. I suspect that for the inductive step I multiply both sides by $(a+b)$, but from there I'm not sure where to go.
So this is what I've got so far doing it this way:
We are given that $$(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k$$ Let $n=1$. Clearly $$(a+b)^1=\begin{pmatrix}1\\0\end{pmatrix}a^1+\begin{pmatrix}1\\1\end{pmatrix}b^1=a+b.$$ Now let us assume that this equality holds for some arbitrary $n>1$, namely for the equality below: $$(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k$$ We must now show that this holds for $n+1$. We do this by multiplying each side of the equality above by $(a+b)$: $$(a+b)(a+b)^n=\sum^{n}_{k=0}\begin{pmatrix}n\\k\end{pmatrix}a^{n-k+1}b^k+\sum^{n}_{k=0}\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^{k+1}$$ Note that $$\sum^{n}_{k=0}\begin{pmatrix}n\\k\end{pmatrix}a^{n-k+1}b^k=\sum^{n+1}_{k=1}\begin{pmatrix}n\\k-1\end{pmatrix}a^{???}b^{???}$$
One can work with $(1+x)^n$ for simplicity, then plug $x=ab^{-1}$; but since you're not doing that, I'll keep $a$ and $b$. Note that if we write $$(a+b)^n=\sum_{k=0}^n \binom nk a^kb^{n-k}$$ using our inductive hypothesis then $$(a+b)^{n+1}=(a+b)(a+b)^n=\sum_{k=0}^n \binom nk a^{k+1}b^{n-k}+\sum_{k=0}^n \binom nk a^{k}b^{n-k+1}$$
Modify the indices so that $\displaystyle \binom nk+\binom{n}{k-1}$ appears. Don't forget to use that $\displaystyle \binom nk=0$ if $k<0$ or $k>n$ to keep $k=0$ and obtain $k=n+1$ "for free". In particular, you can simply change the upper limits of the last term to $n+1$, and take $k\to k-1$ on the first sum, then keep $k=0$ since $\displaystyle \binom n{-1}=0$.
SPOILER We have that
and that