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 [imath]x[/imath], of the binomial expansion of [imath](1+px)^{15}[/imath], where [imath]p[/imath] is a non-zero constant.

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

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

 

[bluefrog]

Where do [imath]\left(11\atop 1\right)[/imath] and [imath]\left(11\atop 2\right)[/imath] 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 [imath]p[/imath], should have read [imath]5(-15p)=105(5/7)^2[/imath], Then when had I substituted [imath]-5/7[/imath] back in, the result would have been positive
Thanks