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

• A

  $\sqrt{2}$
• B

  $2 \sqrt{2}$
• C

  $5 \sqrt{2}$
• Correct answer
  D

  $3 \sqrt{2}$

Solution

No. of terms in $(x+y)^{(2 n-3)} \Rightarrow\left[\begin{array}{c}(2 n-3+1) \\ (2 n-2)\end{array}\right].$ $\therefore$ sum of all coefficients $=2^{2 n-3}$ (Put $\mathrm{x}=\mathrm{y}=1$ ) $\therefore$ Arithmetic mean of all coefficients $\begin{aligned} & =\left(\frac{2^{2 n-3}}{2 n-2}\right)=16 \\ & \Rightarrow 2^{2 n-3}=2^5(n-1) \Rightarrow n=5 \\ & \therefore P\left(2 n-1, n^2-4 n\right)=(9,5) \end{aligned}$
$\begin{aligned} & x+y=8 \\ & \therefore P M=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=\frac{3 \times 2}{\sqrt{2}}=3 \sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'D'.