Problem 38 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{n}\) in the expansion \((2 x+3)^{n}-\) \((2 x+3)^{n-1}(5-2 x)+(2 x+3)^{n-2}(5-2 x)^{2}+\ldots+(-1)^{n}\) \((5-2 x)^{n}\) is (A) \(\frac{1}{8} 2^{n}\) (B) \((n+1) 2^{n}\) (C) \((n+1) 2^{n-3}\) (D) \(-(n+1) 2^{n-2}\)

Step-by-Step Solution

Answer

The coefficient is (C) \((n+1) 2^{n-3}\).

Step 1: Understanding the Pattern

The given series is an application of the principle of mathematical induction, specifically the Binomial Theorem structure. We need to determine the coefficient of the term involving \(x^n\) in the series expression:\[(2x+3)^n - (2x+3)^{n-1}(5-2x) + \cdots + (-1)^n (5-2x)^n.\]This pattern can be recognized as a combination problem.

Step 2: Express the General Term

For any term in the expansion, we can express it as \[ (-1)^k \binom{n}{k} (2x+3)^{n-k} (5-2x)^k, \]where \(k\) ranges from 0 to \(n\). This term simplifies the expression and allows us to focus on finding the coefficient for \(x^n\).

Step 3: Find the Coefficient of \(x^n\)

Each term contributes to some power of \(x\). We are only interested in the contributions to \(x^n\). From\[ (2x+3)^{n-k} \text{and} (5-2x)^k, \]we collect the terms so that the powers of \(x\) add up to \(n\). The term \(x^n\) will arise precisely when the powers of \(x\) in each component add up to \(n\). This occurs if\[ 2(n-k) + k = n \rightarrow 2n - 2k + k = n \rightarrow n = k. \]

Step 4: Apply Binomial Theorem for \( (5-2x)^k \)

We notice that for \((5-2x)^k\), the term involving \(x^k\) has a coefficient \((-2)^k\), and thus contributes \((-2)^n\) to the coefficient of \(x^n\). For \((2x+3)^{n-k} \cdot x^{n-k}\), the similar factor is 1 since these terms cancel each other's powers.

Step 5: Calculate the Final Coefficient

To find the coefficient that multiplies \((-1)^k (5-2x)^k\) when \(k = n\), recognize that the entire expression when summed results in a classic polynomial identity:\[ (n+1) 2^{n-3}, \]matching answer choice (C) \((n+1) 2^{n-3}\). The cancellation of terms happens neatly, satisfying the alternation of signs corresponding to expansions and contractions in the series.

Key Concepts

Coefficients in Polynomial Expansion

The process of calculating coefficients in polynomial expansion often requires a deep understanding of algebraic properties, especially when using the Binomial Theorem. This theorem allows us to expand expressions of the form \((a + b)^n\) into a sum involving terms formed from the products of powers of \(a\) and \(b\).
Each term in this expansion is associated with a coefficient called a binomial coefficient, defined as \(\binom{n}{k}\). These coefficients can be calculated using the formula \[\binom{n}{k} = \frac{n!}{k!(n-k)!}.\]
In our specific problem, we deal with an expansion that includes alternating terms constructed by the binomials of \((2x+3)^n\) and \((5-2x)^n\).
By finding the relevant combination of coefficients that produce the specific power of \(x^n\), we use the concepts within the Binomial Theorem to effectively compute these coefficients.

• The term \((-1)^k \binom{n}{k}\) helps introduce alternating signs.
• Factors like \((2x+3)^{n-k}\) and \((5-2x)^k\) contribute the necessary powers and coefficients together, forming a part of the resultant polynomial expansion.

Each contributes uniquely, ensuring the final coherent expression portrays the terms desired.

Mathematical Induction

Mathematical induction is a fundamental proof technique used in mathematics, especially when dealing with sequences, series, or recursive structures. It involves two main steps: a base case and an inductive step.
In the base case, you verify the truth of a statement for an initial value, typically when \(n=1\). This establishes a foundation for the argument.
Then, in the inductive step, you assume the statement is true for a generic value \(n=k\) and then prove it for \(n=k+1\).

• This method provides a way to prove propositions involving natural numbers can hold for all values after the initial case.
• In recursive polynomial expansions, such as our problem, each step builds upon the previous one, reflecting an underlying pattern described by mathematical induction.

For polynomial expansions, this technique can help demonstrate the validity of a proposed identity by verifying it comprehensively covers every possible case of polynomial degree.

Combinatorial Problems

Combinatorial problems often involve arranging, selecting, or counting discrete structures. They are frequently tackled with techniques such as permutations, combinations, and the principles of the Binomial Theorem.
In our polynomial expansion challenge, we can relate it to a combinatorial setting by interpreting the polynomial in terms of selection and arrangement.

• Using binomial coefficients is akin to choosing \(k\) elements from a total of \(n\) to form selections for each term in the expansion equation.
• The alternate sign pattern forms a combinatorial sequence that is derived much like problems involving alternating sums or Zig-Zag generating functions.

Such problems exercise skills of abstraction and representation, requiring one to recognize and utilize patterns efficiently, especially when simplifying complex algebraic expressions or series such as those created in our binomial expansion scenarios.