Pascal's Triangle and Binomial Theorem

AI Thread Summary
To find the coefficient of x^4y^9 in the expansion of (3x + y)^13, the Binomial Theorem is applied, which states that (x + y)^n can be expressed using binomial coefficients. The coefficient is calculated as 3 * C(13,9), leading to an initial value of 2145. However, corrections were made regarding the exponent of x, resulting in the correct coefficient being 34 * C(13,9), which equals 57,915. The discussion highlights the importance of accurately identifying the relationship between coefficients and exponents in binomial expansions. Understanding these concepts is crucial in discrete mathematics.
SurferStrobe
1. Evaluate the numbers for the coefficient of x4y9 in the expansion of (3x + y)13.

2. The Binomial Theorem states that for every positive integer n,
(x + y)n = C(n,0)xn + C(n,1)xn-1y + ... + C(n,n-1)xyn-1 + C(n,n)yn.

3. I understand that the coefficients can be found from the n row of Pascal's triangle, where n = 13. Using the binomial theorem, my approach (which I'm not sure about) is:

The coefficient is 3 * C(13,9) = 3 * 715 = 2145.

Am I going about this correctly? Sorry if I didn't expand on the proof.

surferstrobe
 
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This is the 5th term from the end (if you start from x^13 then diminish x's power), so is C(13,9) consistent with the formula?

And, you should have 34.
 
Last edited:
I was just going to say, I forgot the exponent of x, so should be 34 * C(13,9).

I used the exponent of y for the combination term, because I knew that

C(13,9) = C(13,13-9) = C(13,4).

Therefore,

34 * C(13,9) = 81 * 715 = 57,915.

Is this better?
 
SurferStrobe said:
I was just going to say, I forgot the exponent of x, so should be 34 * C(13,9).

I used the exponent of y for the combination term, because I knew that

C(13,9) = C(13,13-9) = C(13,4).

Therefore,

34 * C(13,9) = 81 * 715 = 57,915.

Is this better?
In the formula x^k is associated with C(n,k-1) isn't it?
 
EnumaElish,

Thank you for helping me to better understand the relationships between the coefficients, exponents, and expressions in this area of discrete mathemetics!

surferstrobe
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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