Question

Using Binomial theorem, find the correct expansion for the binomial: (x+y)^12

Answer 1

We know that binomial expansion of
\[(a+b)^{n}=^nC_0 a^n+^nC_1a^{n-1}b+^nC_2a^{n-2}b^2.......^nC_{n-1}a^{1}b^{n-1}+^nC_nb^n\]

Can you try now?
here a=x, b=y and n=12
@natnatwebb

Answer 2

*facepalm*

Answer 3

Tell me where you have doubt, I'll try to help

Answer 4

No doubt, disbelief I overthought it that much. One question, what is C?

Answer 5

\[^aC_b= \frac{a!}{b!. (a-b)!}\]
You know factorial ?

Answer 6

Barely. its the end of the year and I'm drowning.

Answer 7

\[n!= 1.2.3.4.5... (n)(n-1)\]
\[1!=1\]
\[2!=1.2=2\]
\[3!=1.2.3=6\]
Can you tell what is 5!?

Answer 8

5!=1.2.3.4.5=120

Answer 9

Right, so if you have
\[^4C_3 \]
Can you find its value?

Answer 10

http://en.m.wikipedia.org/wiki/Binomial_theorem

Answer 11

@natnatwebb ???

Answer 12

I'm sorry, yes lemme work it.

Answer 13

Take your time

Answer 14

it = 4

Answer 15

correct, now you work on the binomial expansion .

Answer 16

okay!

Answer 17

Let me know if you face any difficulty

Answer 18

I feel like there should be a much faster way of working this out.

Answer 19

I think this is the fastest way, you just find the values. Oh yes you should know that
\[^nC_0=^nC_n\]
\[^nC_1=^nC_{n-1}\]
so
\[^{12}C_0=^{12}C_{12}=1\]
\[^{12}C_1=^{12}C_{11}\]
\[^{12}C_2=^{12}C_{10}\]
So you need to find only half of the C's

Answer 20

Okay sweet, I got it. C12=1 C11=12 C10=66

Answer 21

find C9=C3
C8=C4
C7=C5
C6
???

Answer 22

=C6? :D

Answer 23

Its far enough to get the answer I need...

Answer 24

Dude, you were awesome, thanks oodles.

Answer 25

yes, just plugin the no.s and it's done

Answer 26

you're welcome @natnatwebb