# Binomial Theorem - Determine n

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1. Feb 12, 2017

### Schaus

1. The problem statement, all variables and given/known data
The sixth term of the expansion of (x-1/5)n is -1287/(3125)x8. Determine n.

2. Relevant equations
tk+1=nCkan-kbk

3. 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!

2. Feb 12, 2017

### Staff: Mentor

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.

3. Feb 12, 2017

### Schaus

So all I need is the x8 and xn-5?
Then bases are the same so - n-5 = 8 ----> n=3?

4. Feb 12, 2017

### Staff: Mentor

$3-5=8$ ?

5. Feb 12, 2017

### Schaus

Whoops! Sorry I meant to put n = 13!