Question

coefficient of x^2..in the expansion of (1-x)(1+2x)(1-3x)(1+4x)...(1+14x)(1-15x)

Answer 1

this equation will form a polynomial of degree 15
 with roots 1,-1/2,1/3.-1/4,....1/15 
the coefficient will be  consecutive sum of product of roots
 $1\frac{-1}{2} \frac{1}{3}.... \frac{-1}{14}+\frac{-1}{2} \frac{1}{3}.... \frac{1}{15}+1\frac{1}{3}.... \frac{1}{15}+...$
$=1\frac{-1}{2} \frac{1}{3}.... \frac{-1}{14} \frac{1}{15}( \frac{1}{1}+ \frac{1}{ \frac{-1}{2} }+ \frac{1}{ \frac{1}{3} }+...) \\=1\frac{-1}{2} \frac{1}{3}.... \frac{-1}{14} \frac{1}{15}(1-2+3-4+...+15) \\ = \frac{-1}{15!}((1+3+5+...15)-(2+4+..+14))\\= \frac{-1}{15!}(8^2-(7)(8))= \frac{-8}{15!}$
so the coefficient is -8 because coefficient of x¹⁵ is 15!

Answer 2

For any polynomial, ( negative of coefficient of x^2)/(Coefficient of highest power of x)

= sum of product of roots taken 2 at a time.

From above answer sum of product of roots = -8/15!

-b/a=-8/15!  

Coefficient of x^15 = -15!

So, b/15!=-8/15!

So b=-8