For an integer n ≥ 2, if the arithmetic mean of all coefficients in the binomial expansion of (x + y

Q: Wait, why is the sum of coefficients $2^m$?

This is a classic property! In the expansion of $(x+y)^m$, the sum of coefficients is found by setting the variables $x$ and $y$ to 1. Since each term is of the form $C(m, k)x^{m-k}y^k$, setting $x=1$ and $y=1$ makes every term just the coefficient itself. Thus, the sum is $(1+1)^m = 2^m$.

Q: Why is the number of terms $m+1$ and not $m$?

Great question! Think about $(x+y)^2 = x^2 + 2xy + y^2$. The power is 2, but there are 3 terms. Similarly, for $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$, the power is 3, but there are 4 terms. The number of terms is always one more than the exponent $m$.

Q: Can I solve $2^{2n-3} = 32(n-1)$ using logarithms?

While you could technically use logarithms, it would lead to a transcendental equation involving $n$ and $\ln(n)$, which is not easily solvable for integers. In JEE problems, when you see an exponential function equated to a linear one, inspection is almost always the intended, fastest path. Always check small integers first!

Question

For an integer $n \ge 2$, 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:

Options:

(A) $\sqrt{2}$

(B) $2\sqrt{2}$

(C) $5\sqrt{2}$

(D) $3\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the binomial expansion $(x + y)^{2n-3}$ and informed that the arithmetic mean of its coefficients is 16. Our objective is to determine the perpendicular distance of point $P(2n - 1, n^2 - 4n)$ from the line $x + y = 8$.

A fundamental property for any JEE aspirant is that the sum of all coefficients in the expansion of $(x + y)^m$ is $2^m$. This is derived by setting $x = 1$ and $y = 1$.

In our specific case, the exponent is $m = 2n - 3$. Therefore, the sum of the coefficients is:

$$2^{2n-3}$$

### The Counting Trap

To calculate the arithmetic mean, we must divide the sum of the coefficients by the total number of terms. The number of terms in the expansion of $(x + y)^m$ is always $m + 1$.

For our expansion, the number of terms is:

$$(2n - 3) + 1 = 2n - 2$$

This simple addition is a common pitfall in high-pressure exam environments. Always ensure you account for the $+1$ when determining the count of terms.

### The Dance of Algebra

Given that the arithmetic mean is 16, we establish the following equation:

$$\frac{2^{2n-3}}{2n-2} = 16$$

Factoring a 2 out of the denominator, we simplify the expression:

$$\frac{2^{2n-3}}{2(n-1)} = 16 \Rightarrow 2^{2n-4} = 16(n-1)$$

Since $16 = 2^4$, we can rewrite the equation as:

$$2^{2n-4} = 2^4(n-1)$$

By testing values, if we set $n = 5$, the left side becomes $2^{2(5)-4} = 2^6 = 64$. The right side becomes $2^4(5-1) = 16 	imes 4 = 64$. Thus, we have confirmed **$n = 5$**.

### The Geometric Finale

With $n = 5$, we determine the coordinates of point $P(2n - 1, n^2 - 4n)$:

$$x = 2(5) - 1 = 9$$
$$y = 5^2 - 4(5) = 25 - 20 = 5$$

The point is $P(9, 5)$. We now calculate the perpendicular distance $d$ from $P$ to the line $x + y - 8 = 0$ using the formula:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our values:

$$d = \frac{|1(9) + 1(5) - 8|}{\sqrt{1^2 + 1^2}} = \frac{|14 - 8|}{\sqrt{2}} = \frac{6}{\sqrt{2}}$$

Rationalizing the denominator, we obtain:

$$d = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$

The final perpendicular distance is **$3\sqrt{2}$**.

Animated Solution for Mathematics - Binomial Theorem: For an integer $n \geq 2$ , 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 $P (2 n - 1, n^{2} - 4 n)$ from the line $x + y = 8$ is:

Select Answer:

Visualized Solution

Introduction to the Problem

Sum of Binomial Coefficients

Total Number of Terms

Setting up the Arithmetic Mean

Simplifying the Equation

Solving for $n$ by Inspection

Finding Coordinates of Point $P$

Visualizing the Line

Applying the Distance Formula

Final Calculation

Rationalizing the Denominator

Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Counting Trap

The Dance of Algebra

The Geometric Finale

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