binomial expansion question

[santeria13]

the question is , in the binomial expansion of :



, the coefficient of x^2 is 28. Find the value of n.

right so i used the general binomial expansion formula and got that:



my problem is that that would make the formula 8n^2 - 8n -56 = 0 and that would give n two values when it only asks for one?

where have i gone wrong?

 

[pka]

the question is , in the binomial expansion of :

View attachment 3161

, the coefficient of x^2 is 28. Find the value of n.

right so i used the general binomial expansion formula and got that:

View attachment 3162

my problem is that that would make the formula 8n^2 - 8n -56 = 0 and that would give n two values when it only asks for one? where have i gone wrong?

Where did the 4 come from? It should be \(\displaystyle \dbinom{n}{2}=\dfrac{n(n-1)}{2}=28\).

 

[santeria13]

Where did the 4 come from? It should be \(\displaystyle \dbinom{n}{2}=\dfrac{n(n-1)}{2}=28\).

oh yeah haha yeah that was a stupid mistake

Another quick question regarding binomial expansion is what do i do when i have to expand (1-x)^-1/2

what confuses me is that the formula i know always has x as positive,

 

[HallsofIvy]

I assume you mean that n was always positive. Certainly the "basic" binomial expansion has n a positive integer. But the generalized binomial theorem (http://en.wikipedia.org/wiki/Binomial_theorem scroll down to the generalized theorem) allows n to be any number, positive or negative, integer or non-integer.