How is the first expression here equal to the second expression? First statement  $$\dbinom{k}0a^{k+1}+\dbinom{k}{k}b^{k+1}+\sum_{i=1}^k \dbinom{k}{i}a^{k+1-i}b^i+\sum_{j=1}^k \dbinom{k}{j-1}a^{k+1-j}b^j$$
Second statement 
$$a^{k+1}+b^{k+1}+\sum_{i=1}^k \left(\dbinom{k}{i}+\dbinom{k}{i-1}\right)a^{k+1-i}b^i$$
Also, $j = i + 1$.
I don't understand why the first equation is equal to the second one. The terms $a^{k+1}$ and $b^{k+1}$ are obviously transitioned as it is from the first to the second statement. However, it seems that in the second statement, $j$ has been blatantly switched for $i$, which looks illegal to me considered $j = i + 1$. Is this some error or have I been missing something?
Note: I got the above two from https://drive.google.com/folderview?id=0B4Ns8Djs6-3DRDVhSlVjS0lOM28&usp=sharing&tid=0B4Ns8Djs6-3DbnlTalJhZTAwRU0, which is the solution for a problem. 
I don't understand the transition that happens at line 15 and 17 (16 is a blank line).
 A: There is no error. The catch is in the endpoints of the summation: $j$ ranges from $1$ to $k$ but then $i$ ranges also from $1$ to $k$ instead of $0$ to $k-1$ which accounts for the (blatant) change of $j$ to $i$ and not to $i+1$. Specifically $$\sum_{j=1}^k \dbinom{k}{j-1}a^{k+1-j}b^j\overset{i=j-1}=\sum_{i=0}^{k-1} \dbinom{k}{i}a^{k+1-(1+i)}b^{i+1}=\sum_{i=1}^{k}\dbinom{k}{i-1}a^{k+1-i}b^i$$
A: Both $i$ and $j$ are just place-holders.  Expand the sums out, and you find they disappear:
$$\sum_{i=1}^k \dbinom{k}{i}a^{k+1-i}b^i={k\choose1}a^kb^1+{k\choose2}a^{k-1}b^2+{k\choose3}a^{k-2}b^3+...+{k\choose k}a^1b^k\\
\sum_{j=1}^k \dbinom{k}{j-1}a^{k+1-j}b^j={k\choose0}a^kb^1+{k\choose1}a^{k-1}b^2+{k\choose2}a^{k-2}b^3+...+{k\choose k-1}a^1b^k\\
Sum=\left({k\choose1}+{k\choose0}\right)a^kb^1+\left({k\choose2}+{k\choose1}\right)a^{k-1}b^2+...+\left({k\choose k}+{k\choose k-1}\right)a^1b^k
$$
Then you can bring a letter back in, and the $\Sigma$, to make it 
$$\sum_{w=1}^k\left({k\choose w}+{k\choose w-1}\right)a^{k+1-w}b^w$$
