Problem 34 - A Complete Resource Book in Mathematics for JEE Main

Question

Coefficient of \(t^{24}\) in \(\left(1+t^{2}\right)^{12}\left(1+t^{12}\right)\left(1+t^{24}\right)\) is (A) \({ }^{12} C_{6}+3\) (B) \({ }^{12} C_{6}+1\) (C) \({ }^{12} C_{6}\) (D) \({ }^{12} C_{6}+2\)

Step-by-Step Solution

Answer

The coefficient of \(t^{24}\) is \(\binom{12}{6} + 2\). The answer is (D).

Step 1: Expand the Expression

We start by expanding the given expression \( \left(1+t^2\right)^{12} \left(1+t^{12}\right) \left(1+t^{24}\right) \). The target is to find the coefficient of \( t^{24} \).

Step 2: Consider the Exponent Combinations

Consider the terms formed by choosing powers of \( t \) from each factor: \( \left(1+t^2\right)^{12} \), \( 1+t^{12} \), and \( 1+t^{24} \). To obtain \( t^{24} \), we can have combinations such as:- powers coming entirely from \( \left(1+t^2\right)^{12} \) and nothing from \( t^{12} \) or \( t^{24} \).- powers coming partly from \( \left(1+t^2\right)^{12} \) and partly from \( t^{12} \), or- all power from \( t^{24} \).

Step 3: Use Binomial Theorem to Calculate Coefficients

Apply the binomial theorem:1. For \( \left( 1+t^2 \right)^{12} \), consider terms of the form \( t^{2k} \). The coefficient is \( \binom{12}{k} \).2. From \( \left( 1+t^{12} \right) \), select either \( 1 \) (no power addition) or \( t^{12} \).3. From \( \left( 1+t^{24} \right) \), select either \( 1 \) or \( t^{24} \).

Step 4: Calculate Coefficient Combinations

To form \( t^{24} \), consider the cases:- Only from \( \left( 1+t^2 \right)^{12} \): Pick \( t^{24} \) which is achieved by using \( 12 \) terms of \( t^2 \). The coefficient is \( \binom{12}{12} = 1 \).- Combining \( t^{12} \) with \( t^{12} \) from \( \left( 1+t^2 \right)^{12} \): Choose \( t^{12} \) from \( \left( 1+t^{12} \right) \) and \( t^{12} \) which means \( 6 \) terms of \( t^2 \). The coefficient is \( \binom{12}{6} + 0 \).- Finally, use \( t^{24} \) directly from \( (1+t^{24}) \): Coefficient for this part is \( 1 \) (from \( 1 \) in \( \left( 1+t^2 \right)^{12} \), applying both exponents of other terms \( 0 \)).

Step 5: Summing the Coefficients

Add coefficients from each case to find the total coefficient of \( t^{24} \):\[ 1 + \binom{12}{6} + 1 \]

Step 6: Conclusion

The total coefficient is \( \binom{12}{6} + 2 \). Therefore, the correct answer is option D.

Key Concepts

Coefficient Calculation

When it comes to finding the coefficient of a term in a polynomial, the Binomial Theorem comes in handy. In our exercise, we want to find the coefficient of the term \( t^{24} \) in the expression \( \left( 1+t^2 \right)^{12} \left( 1+t^{12} \right) \left( 1+t^{24} \right) \). This involves several steps:

• First, we expand each factor of the polynomial. In this case, the factors are \( \left( 1+t^2 \right)^{12} \), \( 1+t^{12} \), and \( 1+t^{24} \).
• From \( \left( 1+t^2 \right)^{12} \), we use the Binomial Theorem to find terms of the form \( t^{2k} \) where \( k \) is a whole number.

The Binomial Theorem tells us that the general term for \( \left( 1+t^2 \right)^{12} \) is \( \binom{12}{k} t^{2k} \). The challenge is to determine which combinations of terms from all factors give us \( t^{24} \) and what their coefficients are.
To solve this, we compute the coefficients for every valid way to form \( t^{24} \) by strategically selecting terms from each factor and sum them up.

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial expression in expanded form. It breaks down products like the expression \( \left( 1+t^2 \right)^{12} \left( 1+t^{12} \right) \left( 1+t^{24} \right) \) into a sum of monomials. Here is how we handle it:

• \( \left( 1+t^2 \right)^{12} \) expands using the Binomial Theorem to a series of terms with varying powers of \( t \), like \( \binom{12}{0} + \binom{12}{1} t^2 + \binom{12}{2} t^4 + \ldots + \binom{12}{12} t^{24} \).
• Each term in \( \left( 1+t^2 \right)^{12} \) can be multiplied by \(1\) or \(t^{12}\) from \(1+t^{12} \).
• Finally, it can also be multiplied by \(1\) or \(t^{24}\) from \(1+t^{24} \).

Therefore, to form \( t^{24} \):

• You could take \( t^{24} \) directly from \( \left( 1+t^2 \right)^{12} \) without any \( t \)-power from the other factors.
• Combine a product of two lower-degree terms from the factors, like \( t^{12} \) from both \( \left( 1+t^2 \right)^{12} \) and \(1+t^{12} \), resulting in \( t^{24} \).
• Directly take \( t^{24} \) from \(1+t^{24} \), with the rest of the terms contributing \(0\) to the power.

Combinatorial Analysis

Combinatorial analysis helps us understand the different ways terms can combine to achieve a desired power of a variable. In our case, we apply this to obtain \( t^{24} \). The process involves making choices about which terms to expand and multiply together:

• The most intuitive combination for \( t^{24} \) could be to take \( t^{12} \) from \( 1+t^{12} \) and pair it with \( t^{12} \) from terms like \( \binom{12}{6} t^{12} \) from \( \left( 1+t^2 \right)^{12} \).
• Another valid option is simply picking \( t^{24} \) from the very last factor \( 1+t^{24} \), since its coefficient is 1.
• The use of the Binomial Coefficient helps calculate how many such combinations exist, such as choosing 12 terms of \( t^2 \) to reach \( t^{24} \), calculated as \( \binom{12}{12} = 1 \).

Ultimately, we sum up these combinations' coefficients: \( \binom{12}{6} \) for \( t^{12}t^{12} \), and 1 directly from picking \( t^{24} \), providing a full understanding of how the coefficient \( \binom{12}{6} + 2 \) is achieved. This careful analysis ensures we cover all possible term combinations without missing or double-counting any.