The coefficient of x^{15} in the product (1-x | vedclass

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(C) Let $P(x) = \prod_{k=0}^{15} (1-2^k x)$.We want the coefficient of $x^{15}$ in this product.This is a known identity related to the Gaussian binom

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$2^{105}-2^{121}$

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(C) Let $P(x) = \prod_{k=0}^{15} (1-2^k x)$.We want the coefficient of $x^{15}$ in this product.This is a known identity related to the Gaussian binomial coefficient or $q$-binomial theorem,where the coefficient of $x^n$ in $\prod_{k=0}^{n-1} (1-q^k x)$ is $(-1)^n q^{n(n-1)/2}$.Here,$n=15$ and $q=2$.The product is $\prod_{k=0}^{15} (1-2^k x) = (1-x)(1-2x)(1-4x)\ldots(1-2^{15}x)$.The coefficient of $x^{15}$ is $(-1)^{15} 	imes 2^{0+1+2+\ldots+14} = -1 	imes 2^{\frac{14 	imes 15}{2}} = -2^{105}$.However,checking the options provided,there seems to be a discrepancy in the standard form. Given the structure of the product,the coefficient of $x^{15}$ is $(-1)^{15} \sum_{0 \le i_1 Since we are picking $15$ terms out of $16$ available indices $(0, 1, \ldots, 15)$,the only way to pick $15$ is to exclude exactly one index $j \in \{0, 1, \ldots, 15\}$.The coefficient is $(-1)^{15} \sum_{j=0}^{15} \frac{\prod_{k=0}^{15} 2^k}{2^j} = -\frac{2^{\frac{15 	imes 16}{2}}}{\prod_{k=0}^{15} 2^k} 	imes \sum_{j=0}^{15} 2^{-j} = -2^{120} \sum_{j=0}^{15} 2^{-j} = -2^{120} (2 - 2^{-15}) = -2^{121} + 2^{105} = 2^{105} - 2^{121}$.

The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x) \ldots (1-2^{15}x)$ is

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