Consider the data on x taking the values 0, 2, 4, 8, ..., 2^n with frequencies nC0, nC1, nC2, ..., n

Q: Wait, the first term in the numerator is $0 \cdot {}^nC_0$. Does that mean the binomial expansion formula doesn't apply directly?

That is a brilliant observation! You are absolutely right. The standard binomial expansion $(1+x)^n = \sum_{r=0}^n {}^nC_r x^r$ includes the term for $r=0$, which is ${}^nC_0 x^0 = 1$. In our specific problem, the first term of the numerator is $0 \cdot {}^nC_0 = 0$. So, when we use the expansion $(1+2)^n$, we get $1 + \sum_{r=1}^n {}^nC_r 2^r$. Since our numerator is exactly that summation, we simply subtract the $1$ (which is ${}^nC_0$) from $3^n$ to get the correct sum. It doesn't break the formula; it just requires a small adjustment!

Q: Why can't we just use the property of mean for binomial distribution?

The mean of a binomial distribution $B(n, p)$ is $np$. However, this problem is not asking for the mean of a binomial distribution. It is asking for the mean of a *data set* where the frequencies are binomial coefficients. While the coefficients look like a binomial distribution, the values $x_i$ are powers of 2, not the random variable $r$ itself. Always distinguish between the distribution of the variable and the frequency distribution of a data set!

Q: How do I know when to use the binomial expansion $(1+x)^n$ versus its derivative?

Great question! Look at the power of $x$ in your series. If you have terms like ${}^nC_r x^r$, you are looking at the direct expansion of $(1+x)^n$. If you had terms like $r \cdot {}^nC_r x^{r-1}$, that would be the derivative of the expansion. Since our series is $\sum {}^nC_r 2^r$, it is the direct expansion. Always check the power of the variable attached to the binomial coefficient!

Question

Consider the data on $x$ taking the values $0, 2, 4, 8, \dots, 2^n$ with frequencies $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively. If the mean of this data is $\frac{728}{2^n}$ then $n$ is equal to

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Data

Welcome, future engineer. Today, we are not just crunching numbers; we are peeling back the layers of a problem that hides a beautiful, elegant symmetry.

When you look at a data set where values are $0, 2, 4, 8, \dots, 2^n$ and frequencies are binomial coefficients ${}^nC_0, {}^nC_1, \dots, {}^nC_n$, you might feel overwhelmed. But stop. Breathe. Look closer.

These aren't random numbers. They are the building blocks of the Binomial Theorem. Let us embark on this journey to find $n$.

### Phase 1: The Anatomy of the Mean

Every statistical problem has a heartbeat, and for a frequency distribution, that heartbeat is the formula for the mean:

$$\bar{x} = \frac{\sum_{i=0}^n x_i f_i}{\sum_{i=0}^n f_i}$$

This formula is our North Star. We have two distinct tasks: calculate the denominator (the total frequency) and the numerator (the sum of products).

Let us tackle the denominator first. It is the sum of all binomial coefficients: $\sum_{r=0}^n {}^nC_r$.

If you have spent time with Pascal's Triangle, you know this identity by heart: the sum of the $n$-th row is exactly $2^n$. So, our denominator is $2^n$. Simple, clean, and powerful.

### Phase 2: The Binomial Magic

Now, the numerator. This is where the thrill begins. We need to calculate:

$$\sum_{i=0}^n x_i f_i = (0 \cdot {}^nC_0) + (2^1 \cdot {}^nC_1) + (2^2 \cdot {}^nC_2) + \dots + (2^n \cdot {}^nC_n)$$

The first term is zero, which is a gift—it simplifies our life. We are left with $\sum_{i=1}^n 2^i {}^nC_i$.

This looks hauntingly familiar. Recall the standard binomial expansion:

$$(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2 x^2 + \dots + {}^nC_n x^n$$

If we set $x=2$, the right side becomes ${}^nC_0 + {}^nC_1(2) + {}^nC_2(2^2) + \dots + {}^nC_n(2^n)$. This is exactly the series we need, plus that pesky ${}^nC_0$ term!

### Phase 3: The Final Act

We know $(1+2)^n = 3^n$. So, $3^n = {}^nC_0 + \sum_{i=1}^n 2^i {}^nC_i$.

Since ${}^nC_0 = 1$, our numerator sum is simply $3^n - 1$. Now, we bring it all together.

The mean is:

$$\bar{x} = \frac{3^n - 1}{2^n}$$

The problem tells us the mean is $\frac{728}{2^n}$. Equating these, we get:

$$\frac{3^n - 1}{2^n} = \frac{728}{2^n}$$

The denominators cancel out, leaving us with the elegant equation $3^n - 1 = 728$, or $3^n = 729$.

We know that $3^6 = 729$. Thus, **$n=6$**.

You have navigated the complexity and arrived at the truth. Keep this clarity, and you will conquer any problem the JEE throws your way.

Animated Solution for Mathematics - Statistics: Consider the data on $x$ taking the values $0, 2, 4, 8, \dots, 2^{n}$ with frequencies $^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}$ respectively. If the mean of this data is $\frac{728}{2 ^{n}}$ then $n$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Data Set

The Mean Formula

Calculating Total Frequency $\sum f_{i}$

Setting up the Numerator $\sum x_{i} f_{i}$

Identifying the Binomial Series

Substituting $x = 2$ in Binomial Expansion

Simplifying the Numerator

Forming the Mean Equation

Solving for $3^{n}$

Finding the final value of $n$

The Sigma Insight: Measures of Central Tendency (Mean, Median, Mode)

The Hidden Symmetry of Data

Phase 1

Phase 2

Phase 3

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