MHB Solve Binominal Form (4x+3)^n | Binomial Coefficients

[Alexstrasuz1]

In solved binominal form (4x+3)^n has two members x^4 and x^3 whose binomial coefficients are equal.
I'm kinda good in solving binomial coefficient, but I never stumbled to something like this

 

[earboth]

In solved binominal form (4x+3)^n has two members x^4 and x^3 whose binomial coefficients are equal.
I'm kinda good in solving binomial coefficient, but I never stumbled to something like this

Hello,

take Pascal's triangle of binomial coefficient and look (for n > 4)
View attachment 3393

for those neighbouring coefficients which are in the relation 3 to 4.

The first hit is for n = 6.

Expanding $$(4x+3)^6$$ you'll find that the coefficients in question are 34560.

 

[topsquark]

In solved binominal form (4x+3)^n has two members x^4 and x^3 whose binomial coefficients are equal.
I'm kinda good in solving binomial coefficient, but I never stumbled to something like this

The binomial expansion of
[math](4x + 3)^n = \sum_{i = 0}^n {n \choose i} (4x)^i \cdot 3^{n- i}[/math]

So the coefficient of the [math]x^3[/math] term (which implies i = 3) is
[math]{n \choose 3}4^3 \cdot 3^{n - 3}[/math]

and the coefficient of the [math]x^4[/math] term (which implies i + 1 = 3 + 1) is
[math]{n \choose 4} 4^4 \cdot 3^{n - 4}[/math]

Equating these:
[math]{n \choose 3}4^3 \cdot 3^{n - 3} = {n \choose 4} 4^4 \cdot 3^{n - 4}[/math]

[math]3 {n \choose 3} = 4 {n \choose 4}[/math]

[math]3 \cdot \frac{n!}{3! (n - 3)!} = 4 \cdot \frac{n!}{4! (n - 4)!}[/math]

[math]3 \cdot \frac{1}{n - 3} = 1[/math]

[math]3 = n - 3[/math]

[math]n = 6[/math]

And you can now calculate that the coefficient is the same as earboth told you, 34560.

-Dan