Question

Use (x+3y^2)^14 to answer the following questions:

How many terms are there?
Is there one middle term or two middle terms?
Explain.
Which numerical term(s) is (are) the middle term(s)?
Find the middle term(s).
Find the last term.

Answer 1

\[(x+3y ^{2})^{14}\] Is what the equation looks like.

Answer 2

Yes, I know that. It's the rest I am having trouble with.

Answer 3

15 terms, One middle term because the number of terms is odd, the middle term would the be the 8th term. Got that much figured.

Answer 4

\[\Large\rm (x+3y^2)^{14}\]

Recall that we start counting from ZERO,
so yes, while the 8th term is the middle term,
that ends up being the 7th when we count from zero, yah?

So maybe we'll call the first term: \(\Large\rm T_0\)
Then the middle term is: \(\Large\rm T_7\)

We use our binomial theorem and get:\[\Large\rm T_7=\left(\begin{matrix}14 \\ 7\end{matrix}\right)x^{14-7}(3y^2)^7\]

Answer 5

Mmm lemme make sure I'm doing that correctly...

Answer 6

\[\Large\rm (x+3y^2)^{14}=\sum_{k=0}^{14} \left(\begin{matrix}14 \\ k\end{matrix}\right)x^{14-k}(3y^2)^k\]

Answer 7

I have a problem in my notes that has 13 terms and the middle term was the 7th term.

Answer 8

Did it correspond to k=7 though?
Or did it correspond to k=6?
Or not sure?

Answer 9

I'm jus saying that our 8th term should correspond to the value k=7.
I shouldn't had called it the "7th term".

Answer 10

I don't know. She never gave us that formula.

Answer 11

k=0 gives us our 1st term,
k=1 our 2nd term,
and so on...
k=7 our 8th term.

Answer 12

No? Ohhh my bad :o

Answer 13

Well you know what the progression of powers looks like? Or no?

Answer 14

x goes down and y counts up

Answer 15

Here is a simple example:
\[\rm(x+y)^4=\text{_}x^4y^0+\text{_}x^3y^1+\text{_}x^2y^2+\text{_}x^1y^3+\text{_}x^0y^4\]Yes good good.

Answer 16

yeah k=7 gives the 8th term in expansion

Answer 17

So notice in this example, the MIDDLE term is the 3rd, but corresponds to a power of 2 on the y, yes?

Answer 18

In the example I left spaces for the coefficients, I hope that wasn't confusing.
Those come from Pascal's Triangle.
They're not filled in yet.

Answer 19

Ok, so the xy in this problem would be \[x ^{7}y ^{7}\]?

Answer 20

The powers should add up to 14, so yes that looks correct.
Our x was x.
But in your problem your "y", the thing holding the second slot, was actually 3y^2, yes?

Answer 21

\[\Large\rm x^7(3y^2)^7\]

Answer 22

So...\[x ^{7}2187y ^{y}\]?

Answer 23

Or no, y^14

Answer 24

\[\Large\rm \text{__}x^7(3y^2)^7\]Ok good :) But let's also leave a slot for the coefficient in front.

Answer 25

We can either go to Pascal's Triangle and scroll allllll the way down to the 13th row, or we can use the uhhhh.. yah that thing. with the.. factorials, yah?

Answer 26

\[\Large\rm \text{__}\cdot2187x^7y^{14}\]

Answer 27

Factorial or Combination?

Answer 28

Ya :) Combination

Answer 29

\[\left(\begin{matrix}14 \\ 7\end{matrix}\right)=3432\]

Answer 30

\[\Large\rm =3432\cdot2187x^7y^{14}\]Ah! Good good good.

Answer 31

Wait, multiply the 3432 and 2187? wouldn't they just be placed on their respective variables?

Answer 32

Yes they could stay there :)\[\Large\rm =3432x^7(3y^2)^7\]\[\Large\rm =3432x^72187y^{14}\]We have 4 things multiplying together. It's not just a coefficient attached to the x and another attached to the y.
But remember, multiplication is `commutative`, meaning, we can multiply things in any order.

So we can choose to bring both coefficients to the front and combine them.

Answer 33

So I have 7505784x^7y^14?

Answer 34

yayyy good job \c:/

Answer 35

Now we get to find the last term. LOL

Answer 36

So what will the powers be on x and "y"?

Answer 37

On the last term,
your x power will have counted all the way down,
while the y power will have counted all the way up,
yah?

Answer 38

Remember that the coefficient in front is always a 1 for the first and last terms.

Answer 39

No x. Y would be 14

Answer 40

Ahh I gotta go D: my game starting!

Answer 41

Good good, power on x is 0, so x^0 = 1.
No x appears.

Answer 42

Nooooo! You're killing me!!!

Answer 43

lol

Answer 44

See, I don't know where you are but it is 3 am here.

Answer 45

and my y was wrong. It would be (y^2)^14 which would be y^28

Answer 46

"y" should be 14, yes.
Maybe I should stop calling it that :d

Answer 47

Our second term is 14 power,
so \(\Large\rm (3y^2)^{14}\)

Answer 48

I have figured the last term to be 4782969y^28

Answer 49

Boy that's a big number :o
Yah looks correct!

Answer 50

I know. It has taken me a whole page to work it out.

Answer 51

WHUT +_+ I hope you're using a calculator! lol

Answer 52

I had trouble keeping up when taking notes in class so that is why I didn't fully understand it but now I do.

Answer 53

Almost done. Will show you. And yes, calculator.

Answer 54

Barely got it all in the frame. Thought it was going to take two pictures.

Answer 55

You expanded the whole thing? XD
Buhahaha! That's too funny.

Answer 56

Wait wait I have another problem for you to do then:\[\Large\rm (2x+3y)^{73}\]

Answer 57

lol

Answer 58

Yeah....no

Answer 59

And I technically didn't have to expand it but It really helped me to understand how to do it. Now I can finish the assignment.

Answer 60

Ah cool stuff c:

Answer 61

Yeah, I guess. LOL Now don't you want to go help me verify my post on the Discrete board?

Answer 62

Step by step answer
http://www.mathskey.com/question2answer/23988/trig-question