Archit's Challenge 3

this is similar to archit sir, but use of a little calculus and further elucidation. the eqn can be written as $f(x)=\sum_{n=0}^{14} ((14-n)x^n)=14\sum_{n=0}^{14} (x^n)-\sum_{n=0}^{14} (nx^n)$ by gp $\sum_{n=0}^{14} (x^n)=\dfrac{x^{15}-1}{x-1}$ d.w.r.x and multiply by x: $\sum_{n=0}^{14} (nx^n)=\dfrac{15x^{15}}{x-1}-\dfrac{x^{16}-x}{(x-1)^2}$ so the expression is the same thing as $f(x)=14\dfrac{x^{15}-1}{x-1}-\dfrac{15x^{15}}{x-1}+\dfrac{x^{16}-x}{(x-1)^2}$ we see that a is a primitive 15th root of unity. if you dont know much about them, if y is a nth root of unity≠1, then. $y^n=1 \quad or\quad y^{n-1}+y^{n-2}+...+y+1=0$ also, if y is a nth root of unity, so is $y^k$ for ints k knowing this property, $a^{15}=1$ $p(a)=a^{14}+a^{13}+...+a+1=0$ so, f becomes $f(a)=14\dfrac{a^{15}-1}{a-1}-\dfrac{15a^{15}}{a-1}+\dfrac{a^{16}-a}{(a-1)^2}=\dfrac{15}{1-a}$ so we are just looking for $\dfrac{15^{14}}{(1-a)(1-a^2)(1-a^3)(1-a^4)...(1-a^{14})}$ we know that a polynomyial can be expressed as $\prod(argument-roots)$ . since a, $a^2$ .. are roots of p, $\dfrac{15^{14}}{p(1)}=\dfrac{15^{14}}{15}=15^{\Huge{\boxed{13}}}$