91Ó°ÊÓ

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 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 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.