- #1

Schaus

- 118

- 5

## Homework Statement

The sixth term of the expansion of (x-1/5)

^{n}is -1287/(3125)x

^{8}. Determine n.

## Homework Equations

t

_{k+1}=

_{n}C

_{k}a

^{n-k}b

^{k}

## The Attempt at a Solution

t

_{k+1}=

_{n}C

_{k}a

^{n-k}b

^{k}

t

_{5+1}=

_{n}C

_{5}(x)

^{n-5}(-1/5)

^{5}

This is where I'm stuck. Do I sub in -1287/(3125)x

^{8}to = t

_{6}? If so what do I do from here?

-1287/(3125)x

^{8}=

_{n}C

_{5}(x)

^{n-5}(-1/5)

^{5}?

Any help is greatly appreciated!