Question

What is the exponent of e in the 4th term of the binomial expansion?

(e^2 - 4f)^5

a)6
b)2
c)4
d)3

Answer 1

Well, you can use this to find the 4th term in the expansion or the pascal traingle I suppose..
\[\sum_{r=0}^{5}*\left(\begin{matrix}5 \\ r\end{matrix}\right)*(e^2)^{5-r}*(-4f)^{r}\]
\[\sum_{r=0}^{5}*\left(\begin{matrix}5 \\ r\end{matrix}\right)*(e^2)^{5-r}*(-4)^{r}*(f)^{r}\]

Answer 2

\[\sum_{r=0}^{5}*\left(\begin{matrix}5 \\ r\end{matrix}\right)*e^{10-2r}*(-4)^{r}*(f)^{r}\]

Answer 3

If I remember correctly; the 4th term is when r=3

Answer 4

so whats the exponent on the e?

Answer 5

4th term can either be for r=3 or r=2..

Answer 6

woops, on the exponent of \(e\) then it's still \(r=3\) right?

Answer 7

i think so

Answer 8

anybody understand? sorry im so lost

Answer 9

maybe @ParthKohli: have an easier method; this is the only method which I remember

Answer 10

Yeah, I do have one

Answer 11

Well, my method is easy; than the pascals triangle..

Answer 12

If you take the expansion of \(( a + b)^2 \) for example — you see something.
\(\Rightarrow a^2 + 2ab + b^2\)

In the first term, we have 2 as the exponent of a and 0 as the exponent of b.
In the second term, we have 1 - 1.
In the third term, we have 0 - 2.

The exponents always adds up to n.

Answer 13

So you just start with:
5 - 0 <--- first term
4 - 1<---- second term
3 - 2<---- third term
2 - 3<----fourth term

Answer 14

Now \(e\) would look something like \((e^2)^2 = e^4\) right?

Answer 15

@Mimi_x3 What was I called for? lol. To explain this method or not? :P

Answer 16

yep thanks!!!

Answer 17

You're welcome.

Answer 18

To conclude the method that I used:
\[\binom{5}{3} *(e)^{10-2(3)} *(-4)^{3} *f^{3} => -640e^{4}*f^{3}\]
which is more convenient if you want to find the whole term. (: