Question

What is the fifth term of the binomial expression?(x^2+2/8)^8
With explanation please.

Answer 1

Do I just plug in 5 for x?

Answer 2

Is it, 2/8?

Answer 3

I was asking id that was the coefficient, but it's not.

Answer 4

It's to the 8th power? Not too sure what you meant by coefficient.

Answer 5

To generate terms of \(\Large (A+B)^{n}\), use the formula

\[\Large \binom{n}{k}*(A)^{n-k}*(B)^{k}\]

k is the term number that starts with k = 0 and it progresses until it hits n.

In this case, we want the fifth term. So k = 4 (remember, the first term is when k = 0)

Also, we have n = 8, A = x^2, B = 2/8

Answer 6

I don't know how I'd impute the (n base k) in a calculator. I don't have a graphing calculator.

Answer 7

use pascals triangle then

Answer 8

look in the row that starts with 1, 8, ...

the 5th value in that row will be equal to 8 choose 5

Answer 9

oops sry, 8 choose 4

Answer 10

70 is equal to five.

Answer 11

so

\[\Large \binom{n}{k} = \binom{8}{4} = 70\]

Answer 12

So I would put 70 in for x?

Answer 13

that's your coefficient for that formula I posted above

Answer 14

Oh okay, hold on.

Answer 15

Let me see if I got this.

Answer 16

answer is 1792x?

Answer 17

that doesn't sound right

Answer 18

no it's not correct

Answer 19

My options are.
1792x
1792x^4
1120x^4
1120x^3

Answer 20

the original problem is

\[\Large \left(x^2 + \frac{2}{8}\right)^8\]

right?

Answer 21

Correct.

Answer 22

none of those options work, so there has to be a typo

Answer 23

There's the picture @jim_thompson5910

Answer 24

I'm getting \[\Large \frac{35}{128}x^8\] and wolfram alpha confirms it

http://www.wolframalpha.com/input/?i=(x%5E2%2B2%2F8)%5E8&t=crmtb01

Answer 25

Yeah, I don't understand the options either.

Answer 26

which is why I'm thinking there's a typo