Problem 74 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The coefficient of \(\mathrm{x}^{1012}\) in the expansion of \(\left(1+x^{n}+x^{253}\right)^{10}\), (where \(n \leq 22\) is any positive integer), is (a) 1 (b) \({ }^{10} \mathrm{C}_{4}\) (c) \(4 \mathrm{n}\) (d) \({ }^{253} \mathrm{C}_{4}\)

Step-by-Step Solution

Answer

(b) \(^ {10} C_{4}\) is the coefficient.

Step 1: Identify the Problem

We need to find the coefficient of \(x^{1012}\) in the expansion of \((1+x^n+x^{253})^{10}\). The given constraint is \(n \leq 22\).

Step 2: Apply the General Term Formula

In the expansion of \((a+b+c)^m\), the general term is given by \(\binom{m}{i,j,k} a^i b^j c^k\) where \(i+j+k=m\). So for \((1+x^n+x^{253})^{10}\), the general term is \(\binom{10}{i,j,k} x^{jn} x^{253k}\) which simplifies to \(\binom{10}{i,j,k} x^{jn + 253k}\).

Step 3: Set Up Equation for Exponent

We want \(jn + 253k = 1012\). We need to find integer values of \(j\) and \(k\) that satisfy this equation for \(n \leq 22\).

Step 4: Solve for Integer Solutions

For different integer values of \(k\), solve for \(j\): - If \(k = 4\), then \(253(4) = 1012\), making \(j = 0\), since \(1012 = 1012+0\). - Only \((j=0, k=4)\) gives a valid term with \(i+j+k=10\).

Step 5: Calculate the Binomial Coefficient

For \(j = 0\) and \(k = 4\), calculate the contribution from \(j, k\) to the coefficient: \(\binom{10}{0,6,4} = \binom{10}{4} = 210\).

Step 6: Verifying Other Options

Check if other options have a common factor with combinations: - As \(j=0\), we ensure no \(10C4\) equivalent or compounding from \((1, x^n)\): - The obtained needed \(x^{1012}\) is only derived and integer verified for \(k=4\).

Key Concepts

Coefficient of x in polynomials

Let's explore the concept of the coefficient of a particular term in polynomials. When dealing with polynomial expansions, one common task is finding the coefficient of a specific term, such as \(x^{1012}\). In our given exercise, the polynomial is \((1 + x^n + x^{253})^{10}\). Here, understanding how expansion works is crucial.

• Each term in the expansion comes from the original terms \((1, x^n, x^{253})\) raised to a power and multiplied by a binomial coefficient.
• The general term in such an expansion is represented as \(\binom{10}{i,j,k} x^{jn + 253k}\), where \(i+j+k=10\).
• The task is to identify the set of \((i,j,k)\) values that allow \(x^{jn + 253k}\) to become \(x^{1012}\).

Recognizing that the sum of all coefficients is also impacted by the binomial nature (choosing elements from different parts), we solve for specific powers of \(x\), like \(x^{1012}\), to find the desired coefficient.

Integer solutions in equations

The expansion's general term involves finding integer values \(j\) and \(k\) that satisfy the equation \(jn + 253k = 1012\). Here’s how you can approach it:

• Start by fixing integer values for \(k\) and calculate possible \(j\) values.
• In our solution, we set \(k=4\) because multiplying \(253\) by \(4\) equals \(1012\), leaving \(j=0\) since no further multiplication is needed.
• This solution works under the constraint \(n \leq 22\).

Checking integer combinations is key to ensure equations satisfy such polynomial constraints. Calculating for \(j=0, k=4\) is considered by finding that the combination aligns with the constraints necessary for other parts of the equation \(i+j+k=10\).

Binomial coefficients

Binomial coefficients provide the weights for the terms in a polynomial expansion. They play a vital role, especially in combinatorics and in forming coefficients of polynomial expansions like the one in our problem.

• The coefficients are denoted with the notation \(\binom{n}{k}\), read as "n choose k".
• In the case of \((1+x^n+x^{253})^{10}\), each term is multiplied by its respective binomial coefficient \(\binom{10}{i,j,k}\).
• For our identified solution \((j=0, k=4)\), the term simplifies to \(\binom{10}{0,6,4}\), equating to \(\binom{10}{4}\).

This coefficient is calculated as 210, representing the number of ways to choose terms contributing to the power \(x^{1012}\). Understanding binomial coefficients as part of combinatorial counting can help immensely with similar problems.