Binomial Theorem - Determine n

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Homework Help Overview

The problem involves determining the value of n in the context of the binomial theorem, specifically relating to the sixth term of the expansion of (x - 1/5)^n, which is given as -1287/(3125)x^8.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the binomial theorem and the specific term formula. There is a question about substituting the given term into the equation and how to proceed from that point. Some participants explore the relationship between the powers of x in the expansion and the given term.

Discussion Status

There is an ongoing exploration of the relationship between the terms in the binomial expansion and the provided expression. Some participants have offered insights into the calculations and relationships, but there is no explicit consensus on the value of n yet.

Contextual Notes

Participants are working with the constraints of the binomial theorem and the specific term provided, and there seems to be some confusion regarding the calculations and assumptions about the powers of x.

Schaus
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Homework Statement


The sixth term of the expansion of (x-1/5)n is -1287/(3125)x8. Determine n.

Homework Equations


tk+1=nCkan-kbk

The Attempt at a Solution


tk+1=nCkan-kbk
t5+1=nC5(x)n-5(-1/5)5
This is where I'm stuck. Do I sub in -1287/(3125)x8 to = t6? If so what do I do from here?
-1287/(3125)x8 = nC5(x)n-5(-1/5)5?
Any help is greatly appreciated!
 
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You already wrote ##(-\frac{1}{5})^5=-\frac{1}{3125}## and ##x^{n-5}=x^8##, what is ##n## then? ##\binom{n}{5}## should only serve as a control calculation.
 
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So all I need is the x8 and xn-5?
Then bases are the same so - n-5 = 8 ----> n=3?
 
Schaus said:
So all I need is the x8 and xn-5?
Then bases are the same so - n-5 = 8 ----> n=3?
##3-5=8## ?
 
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Whoops! Sorry I meant to put n = 13!
 

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