# Thread: factor np out of binomial summation

1. ## factor np out of binomial summation

I totally screwed this problem up earlier. I understand what needs to be done but it's been 35 years since I studied sequences and series. Could someone help by giving me a hint on how to proceed?

I need to factor out np from the following sigma summation:

$\displaystyle \sum_{x=1}^{n}\left({n\atop x}\right)xp^{x}(1-p)^{n-x}=np(1-p)^{n-1}+\frac{n!}{2!(n-2)!}2p^{2}(1-p)^{n-2}+\ldots+$

Now factoring out the np is not a problem in itself, and I need to change the limits by letting y=x-1 so I can sum over y=0 to y=n-1, but what do I do with the $\displaystyle (1-p)^{n-1}$ in the first term?

Thanks

2. $\displaystyle \sum_{x = 1} ^ n x \binom{n}{x}p^x (1 - p)^{n - x}$

$\displaystyle = np \sum_{x = 1} ^ n \frac{(n-1)!}{(x - 1)! (n-1 - (x - 1))!} p^{x - 1} (1 - p)^{(n - 1) - (x - 1)}$

$\displaystyle = np \sum_{x = 0} ^ {n - 1} \binom{n - 1}{x}p^x (1 - p)^{(n - 1) - x}$

I'll leave some of the details between the equalities, as well as the final step, to you.