Proof of the binomial theorem by induction clarification This is an exercise from Spivak's "Calculus".

3.d. Prove the "binomial theorem": If $a$ and $b$ are any numbers and $n$ is a natural number, then
\begin{align} (a+b)^n &= a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + \dots + {n \choose n-1}ab^{n-1} + b^n \\ &= \sum_{j=0}^n {n \choose j} a ^{n-j}b^j \end{align}

The solution is given as

The binomial theorem is clear for $n=1$. Suppose that
$$(a+b)^n=\sum_{j=0}^n{n\choose j}a^{n-j}b^j$$
Then
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n=(a+b)\sum_{j=0}^n{n\choose j}a^{n-j}b^j\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=0}^n{n\choose j}a^{n-j}b^{j+1}\\
&=\sum_{j=0}^n{n\choose j}a^{n+1-j}b^j+\sum_{j=1}^{n+1}{n\choose j-1}a^{n+1-j}b^j\ \text{(replacing }j \text{ by } j-1 \text)\\
&=\sum_{j=0}^{n+1}{n+1\choose j}a^{n+1-j}b^j
\end{align}

My question is similar to the one asked here, but it is not about the change of indices. What were the steps to combine the two summations at the end?

It is Pascal's rule:
$${n\choose j}+{n\choose j-1}={n+1\choose j}$$
So we have
\begin{align}
(a+b)^{n+1}&=(a+b)(a+b)^n\nonumber \\
&=(a+b)\sum_{j=0}^{n}{n\choose j}a^{n-j}b^j\nonumber \\
&=\sum_{j=0}^{n}{n\choose j}a^{n-j+1}b^j +\sum_{j=0}^{n}{n\choose j}a^{n-j}b^{j+1}\nonumber \\
&={n\choose 0}a^{n+1}+\sum_{j=1}^n{n\choose j}a^{n-j+1}b^j+\sum_{j=1}^{n+1}{n\choose j-1}a^{n-j+1}b^j\nonumber \\
&={n\choose 0}a^{n+1}+{n\choose n}b^{n+1}+\sum_{j=1}^{n}\left[{n\choose j}+{n\choose j-1}\right]a^{n-j+1}b^j\nonumber \\
&={n+1\choose 0}a^{n+1}+{n+1\choose n+1}b^{n+1}+\sum_{j=1}^n{n+1\choose j}a^{n-j+1}b^j\nonumber \\
&=\sum_{j=0}^{n+1}{n+1\choose j}a^{n+1-j}b^j\nonumber
\end{align}
