Binomial Expansion

[arnav_deorukhkar]

Hi sorry, dont take SAT math so I don't know what algebra 2 calc 2 is etc. Anyways, attatched is the question. Any help would be great.

 

[MarkFL]

Hello, and welcome to FMH!

According to the binomial theorem, the \(n\)th term in the expansion is:

[MATH]{12 \choose n}\left(2x^4\right)^{12-n}\left(\frac{x^2}{k}\right)^n[/MATH]
Can you, using the laws of exponents, write this in the form:

[MATH]a(n)x^{b(n)}[/MATH]
where \(a\) is the coefficient and \(b\) the exponent on \(x\), where both \(a\) and \(b\) are functions of \(n\)?

 

[MarkFL]

To follow up, the \(n\)th term may be written as:

[MATH]2^{12-n}k^{-n}{12 \choose n}x^{48-2n}[/MATH]
The term in \(x^{40}\) must have \(n=4\), and the term in \(x^{38}\) must have \(n=5\), and the ratio of the coefficient of the former term to the latter term is 5:

[MATH]\frac{2^{12-4}k^{-4}{12 \choose 4}}{2^{12-5}k^{-5}{12 \choose 5}}=5[/MATH]
[MATH]\frac{2^{8}k^{-4}\dfrac{12!}{4!8!}}{2^{7}k^{-5}\dfrac{12!}{5!7!}}=5[/MATH]
[MATH]\frac{2k\cdot5}{8}=5[/MATH]
[MATH]\frac{k}{4}=1[/MATH]
[MATH]k=4[/MATH]