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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, 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).

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up vote 2 down vote accepted

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 $$\sum_{w=1}^k\left({k\choose w}+{k\choose w-1}\right)a^{k+1-w}b^w$$

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Are you a wizard? That might seem like common sense to you but it just blew my mind in four different directions. – Aayush Agrawal Jan 16 at 15:33

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 $$\sum_{j=1}^k \dbinom{k}{j-1}a^{k+1-j}b^j\overset{i=j-1}=\sum_{i=0}^{k-1} \dbinom{k}{i}a^{k+1-(1+i)}b^{i+1}=\sum_{i=1}^{k}\dbinom{k}{i-1}a^{k+1-i}b^i$$

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How is it legal to replace i with i - 1? – Aayush Agrawal Jan 16 at 15:42
@AayushAgrawal Because I change accordingly the endpoints of summation. Plug in the first and last value for $i$ in the middle sum ($0$ and $k$) and then repeat and compare with the first and last term in the rightmost sum. – Jimmy R. Jan 16 at 15:49
I understand that. The part which i don't understand is why wouldn't this imply that i = i - 1? (A statement which is ridiculous) – Aayush Agrawal Jan 16 at 15:58
Look at a more simple example: $\sum_{i=0}^{2}i=\sum_{i=1}^{3}(i-1)$. Do you agree? Same thing here. – Jimmy R. Jan 16 at 16:05
Hmm, i now get it (And also feel so stupid). Thank you. – Aayush Agrawal Jan 16 at 16:43

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