Problem 17 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If coefficient of \(x^{n}\) in \((1+x)^{101}\left(1-x+x^{2}\right)^{100}\) is nonzero, then \(n\) can not be of the form (A) \(3 t+1\) (B) \(3 t\) (C) \(3 t+2\) (D) \(4 t+1\)

Step-by-Step Solution

Answer

The coefficient of \(x^n\) cannot be zero if \(n\) is of the form \(3t+2\) (Option C).

Step 1: Expand each factor and focus on the powers of x

We start by recognizing that the problem involves finding the coefficient of \(x^n\) in the expansion of \((1+x)^{101}(1-x+x^2)^{100}\). The term \((1+x)^{101}\) expands according to the binomial theorem, while \((1-x+x^2)^{100}\) must be expanded as well. For the latter, notice that it can be expanded as a product of terms from choosing \(1\), \(-x\), or \(x^2\) from each of the 100 factors.

Step 2: Determine possible powers of x from each factor

The term \((1+x)^{101}\) contributes powers of \(x\) from \(x^0\) to \(x^{101}\). For \((1-x+x^2)^{100}\), a term such as \(x^k\) would arise from selecting \(-x\) at \(k\) places, \(x^2\) at \(j\) places, and \(1\) at the remaining \((100-k-j)\) places, such that the total power of \(x\) is \((k+2j)\).

Step 3: Analysis for non-zero coefficients

The problem requires finding when the coefficient cannot be zero. If the power of \(x\), say \(n\), can be expressed as \((m+k+2j)\), it implies non-zero contribution when split across the two binomials. If we analyze divisibility, this sequence depends on having terms whose degree combines to form \(x^n\) in scenarios not allowed by this divisibility.

Step 4: Consider cycles of 3 and mod condition

Since \(n = m + k + 2j\) with \(m\) from \(x^m\) with \(m\in [0,101]\) and \(k + 2j\) from \(-x+x^2\) expansions, we notice by modular arithmetic (mod 3), it is impossible to form \(x^n\) with \(n\equiv 2 \pmod{3}\). Choosing \(-x\) from \(1-x+x^2\) (\(x^1\)) means the power of \(x\) can never satisfy this congruence, making \(n\equiv 2 \pmod{3}\) an invalid form.

Key Concepts

Binomial Theorem

The Binomial Theorem provides a way to expand expressions that are raised to a power, specifically those in the form of \((a+b)^n\). It tells us that such an expression can be expanded into the sum of terms in the form \(\binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) represents the binomial coefficients, also known as "n choose k". These coefficients can be calculated using the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).

The Binomial Theorem is instrumental when dealing with polynomial and series problems in combinatorial mathematics. For instance, in evaluating \((1+x)^{101}\), this theorem assures us that each term of the expansion involves \(x\) raised to a power that ranges from 0 to 101, with corresponding binomial coefficients.

Understanding how to use this theorem allows you to swiftly expand expressions and determine the coefficients of each term, which is essential for solving problems like the initial exercise.

Polynomial Expansion

Polynomial Expansion refers to expressing a polynomial, which can be a product or a composition of terms, as a sum of monomials. For instance, when dealing with \((1-x+x^2)^{100}\), the expansion involves multiplying multiple binomials together to form a long series.

Key steps include determining how individual components contribute to the general term. In the given formula \((1-x+x^2)^n\), each factor can contribute a choice of three terms: \(1\), \(-x\), or \(x^2\). When expanded completely, combinations of these choices from the multiple binomial factors will define the overall expression.

Therefore, the task becomes one of selection: deciding how many times to choose each term within the polynomial factors to reach a specific power of \(x\). This methodical choosing is combinatorial at its heart, as it requires counting the number of ways different selections can be made to give different powers of \(x\). As a result, mastering polynomial expansion is crucial for handling related problems effectively.

Modular Arithmetic

Modular Arithmetic simplifies calculations by allowing us to work with remainders. It's basically a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value, called the modulus.

In the context of our exercise, modular arithmetic helps determine which powers of \(x\) will result in nonzero coefficients. Specifically, when we are evaluating the polynomial \((1+x)^{101}(1-x+x^2)^{100}\), easily handling terms requires us to check possibilities under modulo conditions.

For example, examining if \(n\equiv 2 \pmod{3}\) means inspecting whether any term from the expansion results in the sequence meeting this specific modular condition. The analysis shows that the form \(3t+2\) is invalid, which corroborates that some arrangements of terms will not form valid expressions in context to the modulus checked.

This approach greatly facilitates complex arithmetic and logical deduction in combinatorial problems, providing a neat tool to quickly rule out impossible configurations.