Solve Binomial Theorem Problem: a+2b+4c=10^30

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SUMMARY

The discussion focuses on finding the number of nonnegative integer solutions to the equation a + 2b + 4c = 10^30 using generating functions and the Binomial Theorem. The generating function identified is 1/((1-x)(1-x^2)(1-x^4)), which is simplified through partial fraction decomposition. The process involves expanding each fraction as a formal power series and determining the coefficient of x^m for m = 10^30. A Computer Algebra System (CAS) was utilized to find the coefficient, resulting in 62500000000000000000000000000500000000000000000000000000001.

PREREQUISITES
  • Understanding of generating functions
  • Familiarity with partial fraction decomposition
  • Knowledge of the Binomial Theorem
  • Experience with Computer Algebra Systems (CAS)
NEXT STEPS
  • Learn how to derive generating functions for combinatorial problems
  • Study partial fraction decomposition techniques in detail
  • Explore the application of the Binomial Theorem in series expansions
  • Practice using a Computer Algebra System (CAS) for symbolic computations
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Mathematicians, students studying combinatorics, and anyone interested in solving complex algebraic equations using generating functions and the Binomial Theorem.

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Homework Statement



I am trying to find the number of nonnegative integer solutions to a+2b+4c=10^30. I found a generating function, and need to check the coefficient of 10^30.

Homework Equations



The generating function is 1/((1-x)(1-x^2)(1-x^4)). I found the PFD, which is -1/(8(x-1)^3) + 1/(4(x-1)^2) - 9/(32(x-1)) + 1/(16(1+x)^2) + 5/(32(1+x)) + (1+x)/(8(1+x^2))

The Attempt at a Solution



I need to simplify this into an equation using the Binomial Theorem. In class, we related each individual fraction to a series of some sort, or a combination and then combined them all into one equation, but I do not understand it at all. Please help!
 
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You do know how to use the binomial theorem in this case right? Expand each of those fractions as a formal power series. Since the exponent is negative, they expand according to

[tex](1-x)^{-k} = \displaystyle \sum_{n \geq 0} {{n+k-1} \choose {n}} x^n[/tex]

then in order to find the coefficient of [itex]x^m[/itex] (which I will denote [itex]\left[ x^m \right][/itex]) you need to find [itex]\left[ x^m \right][/itex] for each individual fraction and then add them up. Think of each fraction as a power series that you're adding, then [itex]\left[ x^m \right][/itex] will just be the coefficient when you add like terms.

This is going to be a long and disgusting process, but it's entirely doable.

Edit: I think I made a mistake in my original post. My other way just ended up simplifying to the original problem because the exponent of the power series expansion in each case was only 1. It's best just to do it the long way above.
 
Last edited:
Using a CAS, I found the coefficient of [itex]10^{30}[/itex] to be 62500000000000000000000000000500000000000000000000000000001
 
Last edited:

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