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Studying Spivak's Calculus I came across a relation I find hard to grasp. In particular, I want to understand it without using proofs by induction. So please prove or explain the following relationship by not using induction.

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

Thanks in advance.

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    $\begingroup$ Seems like an index change. $\endgroup$ – Alexey Burdin May 26 '15 at 1:46
  • $\begingroup$ What is your summing index? $\endgroup$ – user137731 May 26 '15 at 1:46
  • $\begingroup$ It looks like $i$ and $j$ ought to be the same variable, yes? Or perhaps, $i$ on the LHS and $j$ on the RHS? $\endgroup$ – zahbaz May 26 '15 at 1:51
  • $\begingroup$ Thanks you. I changed that now. I am sorry about that "spelling mistake". $\endgroup$ – Eugene May 26 '15 at 1:52
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    $\begingroup$ All that has really happened is that term by term the sequences refer to the same thing. On the LHS, the third term being summed is when $j=2$ and refers to $\binom{n}{2}a^{n-2}b^{2+1}$, whereas the third term being summed on the RHS is when $j=3$ and refers to $\binom{n}{3-1}a^{n+1-3}b^3$. In general, you could think of it as changing from summing using $j=0..n$ as summing via $j' = j+1 = 1..(n+1)$. As $j' = j+1$ then you see $j'-1 = j$, so everywhere on the LHS you used to see a $j$, on the RHS you see instead a $j'-1$. After making the change, they opt to just call it $j$. $\endgroup$ – JMoravitz May 26 '15 at 2:02
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The identity you've given appears to be an index shift. Instead of beginning to sum at $i=0$, we wish to begin at $1$. In order to advance the summation index ahead by $1$, we have to take away $1$ from every instance of the index variable inside the summand.

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

The index shift becomes clear if you let $j = i + 1$ and substitute.

$$= \sum_{j=0+1}^{n+1}\binom{n}{j-1}a^{n-(j-1)}b^{(j-1)+1}$$ $$= \sum_{j=1}^{n+1}\binom{n}{j-1}a^{n-j+1}b^{j}$$

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  • $\begingroup$ Thank you but it also changes from n to n+1. $\endgroup$ – Eugene May 26 '15 at 2:05
  • $\begingroup$ Ah, so it does! I'll edit it. That changes in the same way. $i=n \implies j = i + 1 = n+1$. The $n$ in the binomial coefficient does not change, though, as it is unaffected by the reindexing. $\endgroup$ – zahbaz May 26 '15 at 2:08

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