Find the coefficient of x^{64} in the expansi | vedclass

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(B) The given expression is $P(x) = \prod_{k=1}^{10} (x - k)^{k+1}$.The total degree of the polynomial is $\sum_{k=1}^{10} (k+1) = \sum_{k=1}^{10} k +

Vedclass · Accepted Answer

$-440$

Vedclass · Answer

(B) The given expression is $P(x) = \prod_{k=1}^{10} (x - k)^{k+1}$.The total degree of the polynomial is $\sum_{k=1}^{10} (k+1) = \sum_{k=1}^{10} k + \sum_{k=1}^{10} 1 = 55 + 10 = 65$.Let $P(x) = a_{65}x^{65} + a_{64}x^{64} + \dots + a_0$.Since the leading coefficient is $1$,$a_{65} = 1$.The coefficient $a_{64}$ is given by the sum of the roots taken with a negative sign: $a_{64} = -\sum_{k=1}^{10} k(k+1)$.$a_{64} = -\sum_{k=1}^{10} (k^2 + k) = -[\frac{10(11)(21)}{6} + \frac{10(11)}{2}]$.$a_{64} = -[385 + 55] = -440$.

Find the coefficient of $x^{64}$ in the expansion of $P(x) = (x - 1)^2(x - 2)^3(x - 3)^4 \dots (x - 10)^{11}$.

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