Binomial Theorem - Determine n

[Schaus]

Homework Statement

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

Homework Equations

t_k+1=_nC_ka^n-kb^k

The Attempt at a Solution

t_k+1=_nC_ka^n-kb^k
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!

 

[fresh_42]

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.

 

[Schaus]

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

 

[fresh_42]

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

$3-5=8$ ?

 

[Schaus]

Whoops! Sorry I meant to put n = 13!