Find the coefficient of the 1996th term of a product

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Homework Help Overview

The problem involves evaluating the coefficient of the 1996th term in the expansion of a product defined as ##\prod_{n=0}^{1996} (1 + nx^{3^n})##. The context is rooted in combinatorial reasoning and properties of powers of 3.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster discusses the uniqueness of powers of 3 in the expansion and how this affects the number of terms. They explore the significance of selecting terms from the product and the implications of powers of 2 summing to 1996. Some participants affirm the correctness of the original poster's reasoning.

Discussion Status

The discussion appears to be productive, with participants providing feedback on the original poster's thought process and formatting. There is a focus on ensuring clarity in mathematical notation, though no explicit consensus on the correctness of the final coefficient has been reached.

Contextual Notes

Participants note issues with formatting in the original post, specifically regarding superscripts and subscripts, which may affect clarity in the mathematical expressions presented.

QuietMind
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Homework Statement


Let ##\prod_{n=0}^{1996} (1 + nx^{3^n}) = 1 + a_1 x^{k_1} + a_2 x^{k_2} + ... + a_m x^{k^m} ## ,
where ##a_1, a_2, ..., a_m ## are nonzero and ##k_1 < k_2 < ... < k_m ##. Find ##a_{1996}##.
From Art and Craft of Problem Solving, originally from Turkey, 1996

Homework Equations


N/A

The Attempt at a Solution


I think I have a solution, but I'm having difficulty explaining myself. I would like to get an evaluation of this thought process:

I think it is relevant that powers of 3 cannot add up to equal another power of 3 and ##\sum_{k=1}^{n-1} 3^k < 3^n##. This means that when the product is expanded out completely, no two terms will have the same power of x, so no simplification of terms will be possible. Because it is a binomial, this means that the number of terms when expanding out the first k products (ie expand ##\prod_{n=1}^{k} (1 + nx^{3^n}) ##)is equal to twice the number of terms when expanding out the first k-1 products.

We don't need to completely compute out the 1996 factors, as we are essentially being asked to describe the 1996th smallest term (if we compute out all the factors we will have many more than that). These smaller terms would be obtained by taking the ##nx^{k^n}## contribution from some or all of the "smaller" terms and taking the 1 from the "larger" terms.(Because of the properties of the powers of 3 that I stated above, we don't need to consider the factors that contribute too large a power of 3, because the inclusion of their x contribution would instantly push us over the 1996th smallest term. This cutoff happens at n=11) We want to find what sum of powers of 2 would yield 1996, which is ##2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^3 + 2^2 = 1996##.
This tells us that the 10th, 9th, 8th, 7th, 6th, 3rd and 2nd factors contribute their ##nx^{3^n}## term, while the other factors contribute their 1 (so are essentially ignored). The coefficient is then ##10*9*8*7*6*3*2 = 181440##.

Does this seem correct, or can I clarify any steps that are not sufficiently explained? I'm still in the process of learning how to write thorough proofs.

Edit: LaTeX formatting
 
Last edited:
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Looks right to me. Well done.
Some of you superscripts/subscripts are a little awry.
 
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haruspex said:
Looks right to me. Well done.
Some of you superscripts/subscripts are a little awry.

Thank you! I revised the formatting for anyone who happens to come across this later.
 
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QuietMind said:
Thank you! I revised the formatting for anyone who happens to come across this later.
I still see a exponents km and kn that should be km and kn.
 

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