Solving Binomial Expansions: Coefficient of x^k in (2x-1/x)^2007

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SUMMARY

The discussion focuses on determining the coefficient of x^k in the binomial expansion of (2x - 1/x)^2007. The correct approach involves using the binomial theorem, specifically the term C(2007, k) * (2x)^(2007-k) * (-1/x)^k, which simplifies to C(2007, k) * (2)^(2007-k) * x^(2007-2k). The key takeaway is that the power of x in the expansion is x^(2007-2k), which clarifies the relationship between the coefficients and the variable x.

PREREQUISITES
  • Understanding of binomial expansions and the binomial theorem
  • Familiarity with combinatorial notation, specifically binomial coefficients C(n, k)
  • Basic algebraic manipulation of expressions involving exponents
  • Knowledge of polynomial terms and their coefficients
NEXT STEPS
  • Study the binomial theorem in depth, focusing on its applications in polynomial expansions
  • Explore combinatorial mathematics, particularly the properties of binomial coefficients
  • Practice problems involving polynomial coefficients and variable manipulation
  • Investigate advanced topics in algebra, such as generating functions and their relation to binomial expansions
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Mathematics students, educators, and anyone interested in combinatorial algebra and polynomial expansions will benefit from this discussion.

brunie
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Hi,
Im having some troubles with this binomial expansion...

Determine the coefficient of x^k, where k is any integer, in the expansion of (2x - 1/x)^2007.

I figured it would just be
C(2007,k) * (2x)^2007-k * (-1/x)^k
= C(2007,k) * (2)^2007-k * x^2007-k * (-1/x)^k

therefore the coefficient would only be the constant terms with no variables

but when i tried it on a smaller scale (ie small exponent), and factored it out, the equation i found doesn't work

i think it is due to the common variable x

anyone kno what to do?
 
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x doesn't drop out of that expression. The power is x^(2007-2*k).
 
so
(2x)^2007-k isn't equivalent to (2)^2007-k * x^2007-k ??
 
Sure it is. But x^(2007-k)*(1/x)^k=x^(2007-2*k).
 
ok yes that makes sense
thanks for ur help
 

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