Question

Binomial theorem,need help finding the co-efficient!!

Answer 1

Long time but I try..

Put some question so that I can make you understand about it..

Answer 2

\[(1+x^{2})^{12} (1+x^{12}) (1+x^{24})\]

Answer 3

co-eff of x^24 in this

Answer 4

Coefficient of which term??

Answer 5

^^

Answer 6

I think we can start by multiplying last two factors..

Answer 7

okay..

Answer 8

By we, I mean you...

Answer 9

\[(1+x^{12}+x^{24}+x^{36})(1+x^{2})^{12}\]

Answer 10

1^{36} will be discarded since we dont need it,next?

Answer 11

Yeah great..

Answer 12

Yes you can..

Answer 13

then?

Answer 14

Can you expand (1+x^2)^12 ?? Using binomial??

Answer 15

no!

Answer 16

Do you know this formula:

\[T_{r+1} = ^n C _r \cdot (x)^{n-r} \cdot (y)^r\]

Answer 17

yes!

Answer 18

Can you find which term will have x^{24} there by using this formula??

Answer 19

i dont know how to do this part and apply this formula thats it!

Answer 20

n is 12 here..

So we have (1+x^2)^12, x = 1 here and y = x^2 here..

So for x^24, There you have \(y^r\) and you have \(y = x^2\) So:

\[(x^2)^r\]
Now to get 24, r should be 12..

Answer 21

Getting this part??

Answer 22

one minute !

Answer 23

i didnt get it :/

Answer 24

In simple words, there I have written formula for this \((x + y)^{n}\)

But you have : \((1 + x^{2})^{12}\)

So compare them..

Answer 25

By comparing you have : n = 12, x = 1 and y = \(x^2\)

Answer 26

I know this is confusing you. Let me try once again..

Re write the above formula for : \((a + b)^n\)

\[T_{r+1} = ^n C _r \cdot (a)^{n-r} \cdot (b)^r\]

This much getting??

Answer 27

alright!!

Answer 28

But you don't have a and b rather you have \((1 + x^2)\)

So here: a = 1 and b= \(x^2\)

Now getting??

Answer 29

yes!

Answer 30

GOT IT!!

Answer 31

thanks!

Answer 32

Can you go further or not??