Is the following equation true?

Question

I observe that for an odd prime p, $\sum_{i=0}^{p-2} \binom{p-2}{i}$$ (i+1)$ = $2^{p-3}$$p$. Can you help me prove this?

Answer 1

As mentioned in the comments above, this is incorrect for all values of $p$ prime or otherwise (with the exception of $p=0$ and $p=2$).

What is true is that:

$$\sum\limits_{i=0}^{p-2}\binom{p-2}{i}=2^{p-2}$$

This is true simply as a special case of the binomial theorem:

$$(x+y)^n = \sum\limits_{i=0}^n \binom{n}{i}x^iy^{n-i}$$

which in the case of $x=y=1$ gives us

$$2^n = \sum\limits_{i=0}^n\binom{n}{i}$$

In this final line, replacing $n$ by $p-2$ gives the aforementioned result.

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In regards to the question in the edit, this again should not have anything to do with $p$ being a prime or otherwise.

$$\sum\limits_{i=0}^n \binom{n}{i}x^i = (1+x)^n$$

Deriving both sides with respect to $x$ and then multiplying both sides by $x$ we get the identity

$$\sum\limits_{i=0}^n\binom{n}{i}ix^{i}=x(1+x)^{n-1}$$

We have then by plugging in $x=1$ and using $p-2$ instead of $n$:

$$\sum\limits_{i=0}^{p-2}\binom{p-2}{i}(i+1) = \sum\limits_{i=0}^{p-2}\binom{p-2}{i}i + \sum\limits_{i=0}^{p-2}\binom{p-2}{i} = 2^{p-3}+2^{p-2} = 3\cdot 2^{p-3}$$

We have as a result what you noticed is correct in the specific case that $p=3$ and in no other situation.