In the expansion of (x-1)(x-2) ldots (x-18), | vedclass

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(B) The expression is given by the product $P(x) = (x-1)(x-2) \ldots (x-18)$. This is a polynomial of degree $18$. The coefficient of $x^{n-1}$ in the

Vedclass · Accepted Answer

$-171$

Vedclass · Answer

(B) The expression is given by the product $P(x) = (x-1)(x-2) \ldots (x-18)$. This is a polynomial of degree $18$. The coefficient of $x^{n-1}$ in the expansion of $(x-a_1)(x-a_2) \ldots (x-a_n)$ is given by $-(a_1 + a_2 + \ldots + a_n)$. Here, $n = 18$ and the terms are $a_1=1, a_2=2, \ldots, a_{18}=18$. The coefficient of $x^{17}$ is $-(1 + 2 + 3 + \ldots + 18)$. Using the sum formula for the first $n$ natural numbers, $S_n = \frac{n(n+1)}{2}$. $S_{18} = \frac{18 	imes 19}{2} = 9 	imes 19 = 171$. Therefore, the coefficient is $-171$.

In the expansion of $(x-1)(x-2) \ldots (x-18)$, the coefficient of $x^{17}$ is:

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