# Binomial expansion of (1+x)

#### rnck

the coefficient of $$\displaystyle x^3$$ is larger than $$\displaystyle x^2$$ by four times....

i can do it easily by drawing Pascal Pyramid.....is there an equation method? i need scores in exam....

#### MarkFL

We could state:

$$\displaystyle (1+x)^n=\sum_{k=0}^n{n \choose k}x^{k}$$

So, we require:

$$\displaystyle {n \choose 3}=4{n \choose 2}$$

$$\displaystyle \frac{n!}{3!(n-3)!}=4\frac{n!}{2!(n-2)!}$$

Can you finish, since it is for an exam, which hopefully, you are allowed outside help?

#### rnck

long ago finished.....now i am doing practices for sitting november general exam.

because my school starts 2 weeks slower than the others which is end of June....all teachers are rushing syllabus like wild stallion.

i find myself unable to form simple notations.......(if you tell me expand its ok , expand with limitations on power is ok , but now question tells me to reverse binomial process and find back , i don't know.)

#### MarkFL

If you would like to post an example of reversing the expansion, I can give you some tips...

#### rnck

i thought this question that suddenly ask you to find value from within a binomial is already demanding one to think reversely?

#### MarkFL

Okay, I see what you mean. I was thinking you were given an entire expansion and needed to find the factorization, or binomial raised to a power.

The problem you posted may be solved using the binomial theorem and the stated relationship between coefficients of 2 particular terms.

#### rnck

Hello Mark...i am back......a question twisted a little and i am dead.

if $$\displaystyle (a+b)^n$$ is a general form of a binomial expansion.

find the 4th and 5th term of a binomial expansion $$\displaystyle (1+\frac{1}{3}x)^n$$
such that the coefficient of the 4th and 5th term is SAME....

if it is a and b i can find......after all it will be odd term for n such that n is 7 and $$\displaystyle ^{7}C_{3} \ \text{and} \ ^{7}C_{4} \ \text{is the same}$$

1 power how many is always the same is one concern but the 1/3 particularly is affected greatly ........
$$\displaystyle ^{n}C_{3}(1)^{(n-3)}(\frac{1}{3})^{3} \ = \ ^{n}C_{4}(1)^{(n-4)}(\frac{1}{3})^{4}$$

#### MarkFL

You have the right idea:

$$\displaystyle {n \choose 3}\(\frac{1}{3}$$^3={n \choose 4}$$\frac{1}{3}$$^4\)

$$\displaystyle 3{n \choose 3}={n \choose 4}$$

$$\displaystyle 3\cdot\frac{n!}{3!(n-3)!}=\frac{n!}{4!(n-4)!}$$

$$\displaystyle 3\cdot4!(n-4)!=3!(n-3)!$$

$$\displaystyle 12=n-3$$

$$\displaystyle n=15$$