Problem 77 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \((1+x)^{15}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{15} x^{15}\), then the value of \(C_{2}+2 C_{3}+3 C_{4}+\ldots+14 C_{15}\) is (A) 219923 (B) 16789 (C) 219982 (D) none of these

Step-by-Step Solution

Answer

The value does not match any option; answer is (D) none of these.

Step 1: Understanding the Expression

Given the expression \((1+x)^{15}\), it can be expanded using the Binomial Theorem as \(C_{0} + C_{1}x + C_{2}x^2 + \ldots + C_{15}x^{15}\). Here, \(C_{k}\) represents the binomial coefficient \(\binom{15}{k}\).

Step 2: Simplifying the Sum

We are tasked with finding the sum \(C_{2} + 2C_{3} + 3C_{4} + \ldots + 14C_{15}\). To simplify this expression, observe that each term of the form \(kC_{k}\) can be derived from differentiating the generating function \((1+x)^{15}\).

Step 3: Differentiating the Generating Function

Differentiate the generating function \((1+x)^{15}\) with respect to \(x\). This yields \((15) \times (1+x)^{14}\). This differentiation represents the sum \(C_{1} + 2C_{2} + 3C_{3} + \ldots + 15C_{15}\).

Step 4: Evaluating at x=1

Substitute \(x=1\) into the differentiated generating function:\[15 \times (1+1)^{14} = 15 \times 2^{14} = 15 \times 16384 = 245760\]

Step 5: Adjusting the Sum

Notice that the expression we found in Step 4 is \(C_{1} + 2C_{2} + 3C_{3} + \ldots + 15C_{15}\). To find \(C_{2} + 2C_{3} + 3C_{4} + \ldots + 14C_{15}\), subtract \(C_{1}\) from the result in Step 4. Here, \(C_{1} = \binom{15}{1} = 15\).Thus, we have:\[245760 - 15 = 245745\].

Step 6: Comparing Options

After adjusting for \(C_1\), the result found, 245745, is checked against given options. None of the options match the computed value, indicating the answer is (D) none of these.

Key Concepts

Binomial Coefficients

Binomial coefficients are a fundamental part of the binomial theorem. They determine the coefficients of the expanded terms in the expression \((1 + x)^n\). Each coefficient, denoted as \(C_k\), is calculated using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(n!\) represents the factorial of \(n\). Factorials involve multiplying a series of descending natural numbers (e.g., \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)).

In our given exercise, the expression \((1 + x)^{15}\) is expanded as \(C_0 + C_1x + C_2x^2 + \ldots + C_{15}x^{15}\). Each \(C_k\) corresponds to \(\binom{15}{k}\). This tells us the impact of raising \(x\) to different powers in the context of expanding a binomial expression.

• \(C_0 = \binom{15}{0} = 1\)
• \(C_1 = \binom{15}{1} = 15\)
• \(C_2 = \binom{15}{2} = 105\)
• \(C_3 = \binom{15}{3} = 455\)
• ... up to \(C_{15} = \binom{15}{15} = 1\)

Each coefficient gives insight into how larger powers of \(x\) contribute more dramatically to the total sum.

Differentiation

Differentiation is a mathematical method used to determine the rate at which a function changes. It's a core concept in calculus and helps to understand dynamics within mathematical expressions.

In the context of the given exercise, we differentiate the generating function \((1 + x)^{15}\) to simplify the complex summation of terms like \(kC_k\). When we differentiate \((1 + x)^{15}\), we use the power rule of differentiation, which states:\[\frac{d}{dx} x^n = n x^{n-1}\]Applying this rule to \((1 + x)^{15}\), we obtain:\[15(1 + x)^{14}\]

This result, after substituting \(x = 1\), gives:\[15 \times (1 + 1)^{14} = 15 \times 16384 = 245760\]This calculation provides the sum \(C_1 + 2C_2 + 3C_3 + \ldots + 15C_{15}\). By understanding how differentiation simplifies the operation, we can handle the polynomial's terms efficiently.

Generating Function

A generating function is a powerful tool that encodes a sequence of numbers in a formal power series. It's particularly useful in combinatorics to handle sequences in a concise way.

For our problem, the generating function is expressed as \((1 + x)^{15}\). This represents all possible sums of powers for its expansion in one concise equation. The coefficients of each term in the expansion provide insight into how elements of a set are combined.

• \(C_0\) is for choosing zero elements, resulting in \(1\).
• \(C_1\) is the count for selecting one element.
• Continuing this, \(C_k\) gives us how \(k\) elements from a total are combined.

By acting as a template, the generating function helps in producing desired coefficients quickly, especially when combined with calculus such as differentiation. This approach simplify processes like whether an expression can hit a particular sum or condition when altering its components—like finding the sum \(C_2 + 2C_3 + \ldots + 14C_{15}\).
Understanding generating functions empowers you to decipher sequences in elegant and efficient ways.