Problem 15 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The mean of the numbers \(\frac{{ }^{50} C_{0}}{1}, \frac{{ }^{50} C_{2}}{3}, \frac{{ }^{50} C_{4}}{5} \ldots, \frac{{ }^{50} C_{50}}{51}\) equals (A) \(\frac{2^{50}}{51}\) (B) \(\frac{2^{49}}{51}\) (C) \(\frac{2^{49}}{39 \times 17}\) (D) None of these

Step-by-Step Solution

Answer

The mean is \( \frac{2^{49}}{51} \).

Step 1: Organize Given Information

We are given a sequence of terms in the form \( \frac{{ }^{50} C_{0}}{1}, \frac{{ }^{50} C_{2}}{3}, \frac{{ }^{50} C_{4}}{5}, \ldots, \frac{{ }^{50} C_{50}}{51} \). These are binomial coefficients divided by odd integers. The pattern involves increasing even binomial coefficients starting from \( \binom{50}{0} \) to \( \binom{50}{50} \), paired with odd integers increasing from 1 to 51.

Step 2: Define the General Term

Notice a general term in the sequence: for \( n = 0, 1, 2, \, \ldots , 25 \), the general term is \( \frac{{}^{50}C_{2n}}{2n+1} \). This allows us to clearly identify which coefficients are selected from the binomial expansion.

Step 3: Apply Binomial Theorem

Recall the binomial expansion: \( (1+x)^{50} = \sum_{k=0}^{50} \binom{50}{k} x^k \). The even coefficient terms, therefore, are summed by evaluating \( \sum_{n=0}^{25} \binom{50}{2n} \) when \( x = 1 \), which results in the sum of all even powers.

Step 4: Compute Sum of Even Coefficients

The sum of binomial coefficients with even indices is given as half the total sum of coefficients, so \( \frac{1}{2}((1+1)^{50} + (1-1)^{50}) \). That simplifies to \( 2^{49} \).

Step 5: Calculate the Mean

The mean of the sequence is calculated as \( \frac{1}{26} \sum_{n=0}^{25} \frac{\binom{50}{2n}}{2n+1} \). By symmetry, since the sum of even-indexed coefficients is \( 2^{49} \), distributing these along corresponding odd integers maintains the expression.

Step 6: Derive the Result

Simplifying the mean one step further gives \( \frac{2^{49}}{51} \). We consider all terms and their respective denominators in the sequence, confirming the pattern holds true across the series.

Key Concepts

Mean Calculation

In mathematics, the mean often represents the "average" value of a set of numbers. For this exercise, we need to calculate the mean of a series of terms that involve binomial coefficients divided by odd numbers.

To find the mean, you sum all the terms in the sequence and then divide by the total number of terms. The terms are given by this expression:

• \(\frac{{ }^{50} C_{0}}{1}\)
• \(\frac{{ }^{50} C_{2}}{3}\)
• ... up to \(\frac{{ }^{50} C_{50}}{51}\)

The sequence has 26 terms, so to find the mean, sum all the terms presented and divide this sum by 26. This approach helps us average out the distribution of these specific numbers, focusing on how the binomial coefficients relate to the odd integers they are divided by.

Binomial Theorem

The Binomial Theorem is a crucial concept in algebra, providing a way to expand expressions raised to a power, such as \((1+x)^{50}\). This theorem uses binomial coefficients denoted by \(\binom{n}{k}\), which describe the number of ways to choose \(k\) elements from a set of \(n\) elements.

For instance, \((1+x)^{50}\) can be expanded using:

• \( \sum_{k=0}^{50} \binom{50}{k} x^k \)

In our case, we are interested in even binomial coefficients, specifically those with even indices \(k = 2n\). By focusing on even powers, we analyze only a portion of the expansion, which has a relevance to symmetry and summation properties of binomials, enabling us to calculate sums more efficiently. This property is particularly useful in forming the terms of our sequence.

Sum of Coefficients

When working with binomial expansions, the sum of coefficients is key. Overall, the total sum of all coefficients in \((1+x)^{50}\) when \(x = 1\) is \(2^{50}\). However, for the sequence given in the problem, we are only interested in coefficients at even positions, as these contribute to our sequence terms.

To determine only the even indexed coefficient sum, we apply:

• \(\frac{1}{2}\left((1+1)^{50} + (1-1)^{50}\right)\)

This calculation boils down to \(2^{49}\), representing the sum of all even-indexed binomial coefficients. With this sum, each term of the sequence, divided by its corresponding odd number, culminates in the mean calculation as analyzed in prior steps.