What is the fourth term of (x + y)8?

Question

mathslover · Answer

just note the basic formula that is BINOMIAL THEOREM :

anonymous · Answer

It's the 8th row.
$$\left(\begin{matrix}8 \ 4\end{matrix}\right) = \frac{8!}{4!4!}$$Solve this for the coefficient. Then, following the binomial theorem like @mathslover said, place an x^5 and y^3 in front to get the whole term.
$$\frac{8!}{4!4!} \times x^{5}\times y^{3} = ?$$

anonymous · Answer

Or if you have a lot of paper, use Pascal's triangle!

anonymous · Answer

56x4y4??

anonymous · Answer

@sterna 56x4y4??

anonymous · Answer

i mean 56x5y3

anonymous · Answer

8! means 8*7*6*5*4*3*2*1, but you probably know that.  If you divide it by 4!, you just cancel the 4*3*2*1 and you're left with 8*7*6*5.  I think mathcalc is saying to divide by 4! again.  Try that : divide (8*7*6*5) by (4*3*2*1) and see what you get.

turingtest · Answer

$$(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$$where $$\binom nk={n!\over k!(n-k)!}$$the $k+1^{th}$ term is given by $$\binom nk x^{n-k}y^k$$

turingtest · Answer

you want the $4^{th}$ term, so $k=3$

turingtest · Answer

...and $n=8$

anonymous · Answer

However, I get something different using the trusty old Pascal's triangle.

turingtest · Answer

$$\binom 83x^{8-3}y^3={8!\over3!(8-3)!}x^5y^3={8\cdot7\cdot\cancel6\over\cancel{3\cdot2}}x^5y^3=56x^5y^3$$hooray, math works!

turingtest · Answer

so both of the above are wrong :P

anonymous · Answer

Nicely done: that's what Pascal told me as well - so does that help 4everJames?

anonymous · Answer

Oops. Sorry. I knew I forgot something :/

anonymous · Answer

@TuringTest Thanks for clarifying my thoughts :)