Find the coefficient of x^{15} in the product | vedclass

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Question

(A) The given product is $P = (1 - x)(1 - 2x)(1 - 2^2 x) \dots (1 - 2^{15} x)$.This can be written as $P = (-1)^{16} (x - 1)(x - 2^{-1})(x - 2^{-2}) \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given product is $P = (1 - x)(1 - 2x)(1 - 2^2 x) \dots (1 - 2^{15} x)$.This can be written as $P = (-1)^{16} (x - 1)(x - 2^{-1})(x - 2^{-2}) \dots (x - 2^{-15}) 	imes (2^0 \cdot 2^1 \cdot 2^2 \dots 2^{15})$.The sum of the exponents is $0 + 1 + 2 + \dots + 15 = \frac{15 	imes 16}{2} = 120$.So,$P = 2^{120} \prod_{k=0}^{15} (x - 2^{-k})$.The coefficient of $x^{15}$ in the expansion of $\prod_{k=0}^{15} (x - a_k)$ is the negative of the sum of the roots,i.e.,$-\sum_{k=0}^{15} 2^{-k}$.Thus,the coefficient of $x^{15}$ is $2^{120} 	imes (- \sum_{k=0}^{15} \frac{1}{2^k}) = -2^{120} 	imes \frac{1(1 - (1/2)^{16})}{1 - 1/2} = -2^{120} 	imes 2(1 - 2^{-16}) = -2^{121}(1 - 2^{-16}) = -2^{121} + 2^{121-16} = 2^{105} - 2^{121}$.

Find the coefficient of $x^{15}$ in the product $(1 - x) (1 - 2x) (1 - 2^2 x) (1 - 2^3 x) \dots (1 - 2^{15} x)$.

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