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So, I was reading the proof and I stumbled upon this part :

$\sum_{k=1}^{n}\binom{n}{k} a^{n-k+1}*b^k$+ $\sum_{k=0}^{n-1}\binom{n}{k} a^{n-k}*b^{k+1}$=$\sum_{k=1}^{n}\{\binom{n}{k}+\binom{n}{k-1}\} a^{n-k+1}*b^{k}$

I'm not exactly sure of understanding how we obtained this final sum. Is there some property which I'm nt aware of?

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    Your curly braces in the right-hand term disappeared, because MathJax interprets them as separators. I put the backslashes in for you. – 2017-01-23
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    I prefer parentheses but I wasn't sure how you do this, this is why I put curly braces. Thanks again for the help ! – 2017-01-23

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In the second sum on the left, set $j=k+1$, so that $k=j-1$. Then $$ \sum_{k=0}^{n-1}{n\choose k}a^{n-k}b^{k+1}=\sum_{j=1}^n{n\choose j-1}a^{n-(j-1)}b^j=\sum_{j=1}^n{n\choose j-1}a^{n-j+1}b^j$$

Changing the variable of summation back to $k$ and combining the two sums leads to the sum on the right.

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    After doing the variable change j=k+1, can I just replace j by k for the sum on the total right directly, is it legal ? – 2017-01-23
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    Sure, but it can sometimes lead to confusion. – 2017-01-23
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    I have a last question in regards of the binomial coefficeint. Can we factor it out of the sums? I assume this is the reason why we're able to add the two of them in the same sum, it's because we factor them out of the sums, and we are left with the two same sums, so we factor these out again, and we replace the binomial coefficient sum back in the unique sum to get the final answer. Am I getting thsi right ? – 2017-01-23
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    No, the binomial coefficients cannot be factored out of the sum because they also depend on the variable of summation. Instead the two sums are simply combined into one sum. – 2017-01-23
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    As long as the other stuff like index and a,b factors are the same in both sums, then I simply can add the coefficients without problem, then ? I just want to be sure, it's the first time i deall with this – 2017-01-23
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    Yes: $\sum_{k=1}^na_k+\sum_{k=1}^nb_k=\sum_{k=1}^n(a_k+b_k)$. – 2017-01-23