For an integer n geq 2,if the arithmetic mean | vedclass

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(D) The number of terms in the expansion of $(x+y)^{2n-3}$ is $(2n-3+1) = 2n-2$.The sum of all coefficients is obtained by setting $x=1$ and $y=1$,whi

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$3\sqrt{2}$

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(D) The number of terms in the expansion of $(x+y)^{2n-3}$ is $(2n-3+1) = 2n-2$.The sum of all coefficients is obtained by setting $x=1$ and $y=1$,which gives $2^{2n-3}$.The arithmetic mean of all coefficients is $\frac{2^{2n-3}}{2n-2} = 16$.This simplifies to $2^{2n-3} = 16(2n-2) = 2^4 	imes 2(n-1) = 2^5(n-1)$.For $n=5$,$2^{2(5)-3} = 2^7 = 128$ and $2^5(5-1) = 32 	imes 4 = 128$. Thus,$n=5$.The point $P$ is $(2(5)-1, 5^2-4(5)) = (9, 5)$.The distance of $P(9, 5)$ from the line $x+y-8=0$ is given by $d = \frac{|9+5-8|}{\sqrt{1^2+1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$.

For an integer $n \geq 2$,if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2n-3}$ is $16$,then the distance of the point $P(2n-1, n^2-4n)$ from the line $x+y=8$ is:

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