The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is .... Problem : 
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is 
(a) $2^{105}-2^{121}$ 
(b)  $2^{121}-2^{105}$ 
(c) $2^{120}-2^{104}$ 
(d) $2^{110}-2^{108}$
My approach : 
Tried to find pattern if any in the given expansion, so expanded first three terms : 
$(1-x)(1-2x)(1-2^2x) = (1-3x+2x^2)(1-4x) = (1-3x+2x^2 -4x+12x^2 -8x^3)$
$= (1-7x+14x^2 -8x^3)$
But didn't find any pattern of this series please suggest how to proceed will be of great help thanks. 
 A: Given $$(1-x)(1-2x)(1-2^2x)...........(1-2^{15}x) = 2^{0+1+2+3+....+15}x^{16}\cdot \left(\frac{1}{x}-1\right)\left(\frac{1}{2x}-1\right).......\left(\frac{1}{2^{15}x}-1\right)$$
So $$=2^{120}x^{16}\left[\left(1-\frac{1}{x}\right)\cdot \left(1-\frac{1}{2x}\right)\cdot \left(1-\frac{1}{2^2x}\right)..........\left(1-\frac{1}{2^{15}x}\right)\right]$$
So $$=2^{120}x^{16}\left[1-\frac{1}{x}\left(1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{15}}\right)+.....\right]$$
So Coefficient of $x^{15} $ in above expression
$$=-2^{120}\left(1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{15}}\right) = -2^{120}\left[1-\frac{1}{2^{16}}\right]\cdot \frac{1}{1-\frac{1}{2}}$$
So we get $$ = -2^{121}\left[1-\frac{1}{2^{16}}\right]=-2^{121}+2^{105}=2^{105}-2^{121}$$
A: I'd rescale as $$2^0\cdot2^1\cdot2^2\cdots2^{15}\left[(1-x)\left(\frac12-x\right)\left(\frac14-x\right)\cdots\left(\frac1{2^{15}}-x\right)\right]$$ The coefficient of $x^{15}$ in the polynomial within the brackets is the negative of the sum of the roots (since there are $16$ linear factors). So the answer is $$-2^0\cdot2^1\cdot2^2\cdots2^{15}\left[1+\frac12+\frac14+\cdots+\frac1{2^{15}}\right]=-2^{120}\left[2-\frac{1}{2^{15}}\right]=2^{105}-2^{121}$$
