Problem 6 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Find \(a\) if the 17 th and 18 th terms of the expansion of \((2+a)^{50}\) are equal.

Step-by-Step Solution

Answer

The value of \(a\) is 1.

Step 1: Understand the Binomial Theorem

The binomial theorem states that for any positive integer \(n\), \[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k.\] This formula helps us find the terms in the expansion. You need to identify the specific terms you are interested in, which are the 17th and 18th terms in this problem.

Step 2: Write the General Term of the Binomial Expansion

The general term in a binomial expansion can be written as \[T_k = \binom{n}{k-1} x^{n-(k-1)} y^{k-1}.\] Here, \(n = 50\), \(x = 2\), and \(y = a\). To find the 17th term, substitute \(k=17\) and for the 18th term, substitute \(k=18\).

Step 3: Derive the 17th and 18th Term

For the 17th term \(T_{17}\):\[T_{17} = \binom{50}{16} (2)^{50-16} a^{16}.\]For the 18th term \(T_{18}\):\[T_{18} = \binom{50}{17} (2)^{50-17} a^{17}.\]

Step 4: Set 17th and 18th Terms Equal

Since the 17th and 18th terms are equal:\[\binom{50}{16} (2)^{34} a^{16} = \binom{50}{17} (2)^{33} a^{17}.\] Divide both sides by \(T_{17}\) to simplify the equation.

Step 5: Simplify the Equation

When simplified, it becomes:\[\frac{\binom{50}{16}}{\binom{50}{17}} \cdot 2 = a.\]Recall that \(\frac{\binom{50}{16}}{\binom{50}{17}} = \frac{17}{34}\), because of the relationship between binomial coefficients.

Step 6: Solve for 'a'

Substituting \(\frac{17}{34}\) into the simplified equation:\[\frac{17}{34} \cdot 2 = a.\]Simplifying gives \(a = 1\).

Key Concepts

Binomial Expansion

When dealing with expressions like \((x + y)^n\),binomial expansion allows us to break down these expressions into manageable terms. This is made possible through the binomial theorem. The theorem states that:\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]This formula is powerful as it guides us in finding each term in the expansion. For instance, if you have \((2+a)^{50}\), you can expand it to find specific terms like the 17th and 18th terms. Understanding how to use this formula with specific values for \(x\), \(y\), and \(n\) is the key to mastering binomial expansion.

• The coefficients \(\binom{n}{k}\) are known as the **binomial coefficients**, which play a vital role in determining the weight of each term.
• This method allows for quick and efficient evaluation of any individual term in a much larger polynomial expression.

Binomial Coefficients

A key component of the binomial expansion is the binomial coefficients, denoted by \(\binom{n}{k}\). These coefficients are determined by the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(!\) denotes factorial, meaning you multiply the given number down to 1. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Binomial coefficients tell us how many ways we can choose \(k\)elements from a set of \(n\)elements, and they give each term in the expansion its respective "weight".

• For the 17th term of \((2+a)^{50},\)the binomial coefficient would be \(\binom{50}{16}\), followed by \(\binom{50}{17}\) for the 18th term.
• Interestingly, binomial coefficients relate to Pascal’s Triangle, which visually represents them. Each entry is the sum of the two directly above it.

General Term

To solve problems involving specific terms in a binomial expansion, we use the concept of the **general term**. This is represented as:\[T_k = \binom{n}{k-1} x^{n-(k-1)} y^{k-1}\]The formula provides a way to calculate any desired term in the expansion without expanding the entire polynomial.
For example:

• To find the 17th term \((T_{17})\) for \((2+a)^{50},\) choose \(k = 17\):\[T_{17} = \binom{50}{16} (2)^{50-16} a^{16}\]
• Similarly, the 18th term \((T_{18})\) is:\[T_{18} = \binom{50}{17} (2)^{50-17} a^{17}\]

In this example, setting these terms equal to each other allowed us to solve for the unknown variable \(a\).Understanding how to carefully substitute and solve the general term formula can provide solutions to complex algebraic problems.