Calculate x^33 Coefficient: Binomial Theorem

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Homework Help Overview

The problem involves using the binomial theorem to find the coefficient of \(x^{33}\) in the expansion of \((\frac{1}{4}-2x^3)^{17}\). The subject area pertains to combinatorics and polynomial expansions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the powers of \(x\) in the expansion and how they relate to the coefficient of \(x^{33}\). There are attempts to clarify how many times \(x^3\) must be used to achieve \(x^{33}\) and the implications of that on the binomial coefficients.

Discussion Status

The discussion is ongoing with various interpretations of how to approach the problem. Some participants are exploring the necessary powers and coefficients, while others are questioning the setup and the sequence of terms in the expansion.

Contextual Notes

There appears to be some confusion regarding the powers of \(x\) and the coefficients involved, particularly about the largest power and how it relates to the binomial coefficients. The original poster and others are considering the implications of these factors in their calculations.

morbello
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use the binomial theorem to determine the coefficient of x^33 in the expansion of ([tex]\frac{1}{4}[/tex]-2x^3)^17

ive played around with it and come up with 33^C_17
as a coefficient.am i right in saying that is all the question asks

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The Attempt at a Solution




 
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You have to find x^33 so the x term should be raised to 11. Cause your x term is x^3 and X^3 power 11 is x^33
 
so is that like 2-11^C_1(x)3 +11^C_2(x)^4 +11^C_3(x)^5 and so on.
 
it would x^3, x^6,x^9 so on
 
in my book it has the largest power on the ^C of C and then the other lower case, is powered up to the sequence it is. in the X^ 3 so the x^3 then x^6 x^9 .what would the largest be the 33 or the 11.

i was thinking the sequence would start at the x^3
 

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