The coefficient of x^{49} in the expansion of | vedclass

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Question

(A) Let $P(x) = (x - 1)(x - \frac{1}{2})(x - \frac{1}{2^2}) \dots (x - \frac{1}{2^{49}})$.This is a polynomial of degree $50$.The coefficient of $x^{5

Vedclass · Accepted Answer

$-2(1 - \frac{1}{2^{50}})$

Vedclass · Answer

(A) Let $P(x) = (x - 1)(x - \frac{1}{2})(x - \frac{1}{2^2}) \dots (x - \frac{1}{2^{49}})$.This is a polynomial of degree $50$.The coefficient of $x^{50}$ is $1$.The coefficient of $x^{49}$ is the sum of the constant terms of each factor multiplied by $-1$,which is $-(1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{49}})$.This is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 50$ terms.The sum $S_{50} = \frac{1(1 - (1/2)^{50})}{1 - 1/2} = 2(1 - \frac{1}{2^{50}})$.Therefore,the coefficient of $x^{49}$ is $-2(1 - \frac{1}{2^{50}})$.

The coefficient of $x^{49}$ in the expansion of $(x - 1)(x - \frac{1}{2})(x - \frac{1}{2^2}) \dots (x - \frac{1}{2^{49}})$ is equal to

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