Binomial expansion of (1+x)

Apr 2010
128
0
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....
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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?
 
Apr 2010
128
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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.)
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
If you would like to post an example of reversing the expansion, I can give you some tips...
 
Apr 2010
128
0
i thought this question that suddenly ask you to find value from within a binomial is already demanding one to think reversely?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
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.
 
Apr 2010
128
0
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 \(\displaystyle ^{7}C_{3} \ \text{and} \ ^{7}C_{4} \ \text{is the same}\)

textbook answer : n=15

1 power how many is always the same is one concern but the 1/3 particularly is affected greatly ........
\(\displaystyle ^{n}C_{3}(1)^{(n-3)}(\frac{1}{3})^{3} \ = \ ^{n}C_{4}(1)^{(n-4)}(\frac{1}{3})^{4}\)
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
You have the right idea:

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

\(\displaystyle 3{n \choose 3}={n \choose 4}\)

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

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

\(\displaystyle 12=n-3\)

\(\displaystyle n=15\)