Binomial expansion problem: first three terms of (1 + px)^15

[bluefrog]

I attempted to solve the following problem:

a) Write down the first three terms, in ascending orders of $x$, of the binomial expansion of $(1+px)^{15}$, where $p$ is a non-zero constant.

b) Given that, in the expansion of $(1+px)^{15}$ , the coefficient of $x$ is $(-q)$ and the coefficient of $x^2$ is $5q$, find the value of $p$ and the value of $q$.

I went about attempting to solve as follows:
a) $$1^{15} +1^{14} {11 \choose 1}(px) + 1^{13} {11 \choose 2}(px)^2 = 1+15px +105p^2x^2$$
b) $$5(15p) = -(105p^2) \\ \Rightarrow -\frac{5}{7}=p, \text{ and } \\ 5q=-105(-\frac{5}{7})^2 \\ q=-\frac{75}{7}$$
The book does not show how the problem is solved, it simply shows the values of $p$ and $q$, as follows:



So the question I have is , how is the sign for part b different to the sign I got, can anybody suggest where I went wrong ?

Thanks

 

[blamocur]

Where do $\left(11\atop 1\right)$ and $\left(11\atop 2\right)$ come from? How did you get 105 ?

If p=-5/7 and q = -15p then what do you get for 'q' ?

 

[bluefrog]

apologies, typo, should read $1^{15}+1^{14}{15 \choose 1}(px) + 1^{13}{15 \choose 2}(px)^2$

 

[bluefrog]

Where do $\left(11\atop 1\right)$ and $\left(11\atop 2\right)$ come from? How did you get 105 ?

If p=-5/7 and q = -15p then what do you get for 'q' ?

so instead of-15p, I had 15p, that was my mistake.

so my equation to solve $p$, should have read $5(-15p)=105(5/7)^2$, Then when had I substituted $-5/7$ back in, the result would have been positive
Thanks