Problem 72 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the number of terms in the expansion of \(\left(1-\frac{2}{x}+\frac{4}{x^{2}}\right)^{n}\), \(\mathrm{x} \neq 0\), is 28 , then the sum of the coefficients of all the terms in this expansion, is: (a) 243 (b) 729 (c) 64 (d) 2187

Step-by-Step Solution

Answer

(b) 729

Step 1: Identifying the General Term

The expression given is \(\left(1 - \frac{2}{x} + \frac{4}{x^2}\right)^n\). The general term in the expansion of \((a + b + c)^n\) is given by \(T_{r} = \frac{n!}{p!q!r!}a^p b^q c^r\), where \(p+q+r = n\). In our case, \(a = 1\), \(b = -\frac{2}{x}\), and \(c = \frac{4}{x^2}\).

Step 2: Setting Condition for Number of Terms

The number of terms in the expansion is given by the number of non-negative integer solutions to \(p + q + r = n\). The formula for this is \(\binom{n+2}{2} = 28\).

Step 3: Solving for n

From \(\binom{n+2}{2} = 28\), we solve:\[\frac{(n+2)(n+1)}{2} = 28\]\[(n+2)(n+1) = 56\]Finding \(n\) gives a quadratic equation:\[n^2 + 3n + 2 - 56 = 0\]\[n^2 + 3n - 54 = 0\]Which can be solved using the quadratic formula or factoring. The factors are \((n+9)(n-6) = 0\), so \(n = 6\).

Step 4: Finding the Sum of Coefficients

The sum of the coefficients of the polynomial \((a + b + c)^n\) can be found by evaluating the expansion at \(x = 1\). Substitute \(x = 1\) in the original expression:\[\left(1 - \frac{2}{1} + \frac{4}{1^2}\right)^6 = (1 - 2 + 4)^6 = 3^6\].

Step 5: Calculating Final Answer

Calculate \(3^6\):\[3^6 = 729\]Therefore, the sum of the coefficients of the expansion is 729.

Key Concepts

Polynomial expansion

Polynomial expansion is a way to express a polynomial raised to a power, in a form that reveals each term separately. In our particular problem, we are expanding \((1 - \frac{2}{x} + \frac{4}{x^{2}})^{n}\).

• This expression has three terms: \(a = 1\), \(b = -\frac{2}{x}\), and \(c = \frac{4}{x^2}\).
• The expansion process involves multiplying and adding different combinations of these terms, raising each to a power as dictated by the exponent \(n\).
• The binomial theorem helps determine each term by using the formula \(T_r = \frac{n!}{p!q!r!} a^p b^q c^r\) where \(p + q + r = n\).

Polynomial expansion allows us to probe into the expression's structure and examine its coefficients and terms, vital for solving the given exercise.

Quadratic equation

A quadratic equation is a polynomial equation of degree two. In most standard forms, it looks like \(ax^2 + bx + c = 0\). For our exercise, while determining the integer \(n\), we derived a quadratic equation: \(n^2 + 3n - 54 = 0\). This form arises from simplifying the expression \((n+2)(n+1) = 56\), a condition provided by the number of terms in the expansion.Solving this quadratic gives insights into the possible values \(n\) can take. We can solve it by:

• Factoring \((n+9)(n-6) = 0\) yields \(n = -9\) or \(n = 6\).
• We discard \(n = -9\) because \(n\) must be a positive integer.

Hence, \(n = 6\), offering a path forward to solving the problem.

Sum of coefficients

The sum of coefficients of a polynomial expansion offers a quick way to understand the basic features of the polynomial without focusing on individual terms. To find the sum of coefficients:Substitute each variable with 1 in the polynomial. So, for \((1 - \frac{2}{x} + \frac{4}{x^{2}})^{6}\), set \(x = 1\).

• This simplifies the expression as: \(\left(1 - 2 + 4\right)^6\).
• The result is \(3^6\) since \(1 - 2 + 4 = 3\).

Calculate \(3^6\) to find the sum of coefficients:- \(3^6 = 729\).This shortcut is powerful when working with expanded polynomials, providing us the sum directly without detailing each term.

Factorial notation

Factorial notation, symbolized by \(n!\), represents the product of all positive integers up to \(n\). It's a fundamental concept when working with permutations and combinations, as factorials count the number of ways to arrange objects. In our exercise, factorial notation surfaced in determining the general term of the polynomial expansion. The formula \(\frac{n!}{p!q!r!}\) assures the proper calculation of each term by accounting for repetitions of each sub-term in the expansion.

• For example, in \(\left(1 - \frac{2}{x} + \frac{4}{x^{2}}\right)^{n}\), each term's coefficient relies on using \(n!\) to comb through all arrangements of \(a^p b^q c^r\) under the condition \(p+q+r=n\).
• Factorial notation simplifies the task of determining these coefficients across various expansions, making the solution neat and mathematically elegant.

Understanding factorials is key as it underpins many combinatorial logic tasks, exemplified in mastering polynomial expansions.