Question

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:

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $5\sqrt{2}$
• D. $3\sqrt{2}$

Hint:

The arithmetic mean of coefficients in a binomial expansion can be used to find the value of $n$.

Answer 1

The Correct Option is D

1. Determine the number of terms in $(x + y)^{2n-3}$: \[ \text{Number of terms} = 2n - 2 \]
2. Sum of all coefficients: \[ \text{Sum of coefficients} = 2^{2n-3} \]
3. Arithmetic mean of all coefficients: \[ \text{Arithmetic mean} = \frac{2^{2n-3}}{2n-2} = 16 \] \[ 2^{2n-3} = 16(2n-2) \] \[ 2^{2n-3} = 2^4(n-1) \] \[ 2n-3 = 4 \implies n = 5 \]
4. Determine the point $P$: \[ P(2n-1, n^2-4n) = P(9, 5) \] 5. Calculate the distance from the line $x + y = 8$: \[ \text{Distance} = \left| \frac{9 + 5 - 8}{\sqrt{2}} \right| = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] Therefore, the correct answer is (4) $3\sqrt{2}$.

Answer 2

We need to find the distance of the point \(P(2n-1,\,n^2-4n)\) from the line \(x+y=8\), given that the arithmetic mean of all coefficients in the expansion \((x+y)^{2n-3}\) is \(16\).

Concept Used:

For \((x+y)^m\), the sum of coefficients is \(2^m\) and the number of coefficients is \(m+1\). Hence the arithmetic mean of coefficients is

\[ \text{Mean}=\frac{2^m}{m+1}. \]

The perpendicular distance from a point \((x_0,y_0)\) to a line \(ax+by+c=0\) is

\[ \text{dist}=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. \]

Step-by-Step Solution:

Step 1: Set up the mean-of-coefficients condition with \(m=2n-3\):

\[ \frac{2^{\,2n-3}}{(2n-3)+1}=\frac{2^{\,2n-3}}{2n-2}=16=2^4. \]

Step 2: Solve for \(n\):

\[ \frac{2^{\,2n-3}}{2n-2}=2^4 \;\;\Longrightarrow\;\; 2^{\,2n-3-4}=2^{\,2n-7}=2n-2. \]

We need an integer \(n\ge2\) satisfying \(2^{\,2n-7}=2n-2\). Checking integers gives \(n=5\).

Step 3: Find the coordinates of \(P\) for \(n=5\):

\[ P(2n-1,\;n^2-4n)=(2\cdot5-1,\;5^2-4\cdot5)=(9,\;5). \]

Step 4: Compute the distance from \(P(9,5)\) to the line \(x+y=8\):

\[ \text{Write the line as } x+y-8=0 \Rightarrow a=1,\;b=1,\;c=-8. \] \[ \text{dist}=\frac{|1\cdot9+1\cdot5-8|}{\sqrt{1^2+1^2}} =\frac{|14-8|}{\sqrt{2}} =\frac{6}{\sqrt{2}} =3\sqrt{2}. \]

Final Computation & Result

The required distance is \(3\sqrt{2}\) units.