Binomial in ratio

[IloveManUtd]

In the expansion of (3+4x)[sup:33uo6uz1]n[/sup:33uo6uz1], the coefficients of x[sup:33uo6uz1]4[/sup:33uo6uz1] and x[sup:33uo6uz1]5[/sup:33uo6uz1] are in the ratio 5:16. Find the value of n.

Do I do like this: [[n(n-1)(n-2)(n-3)(n-4)]/(5!)] / [[n(n-1)(n-2)(n-3)] /(4!)] = 16/5 ? If yes then what next? I'm stuck. Thx.

 

[galactus]

Let me see. It is mighty early.

But, if we write:

\(\displaystyle \frac{\binom{n}{4}\cdot 3^{n-4}(4)^{4}}{\binom{n}{5}\cdot 3^{n-5}(4)^{5}}=\frac{15}{4(n-4)}\)

This reduces to:

\(\displaystyle \frac{15}{4(n-4)}=\frac{5}{16}\)

Solve for n.

 

[soroban]

Hello, IloveManUtd!

\(\displaystyle \text{In the expansion of }(3+4x)^n,\,\text{ the coefficients of }x^4\text{ and }x^5\text{ are in the ratio }5:16.\)

\(\displaystyle \text{Find the value of }n.\)

\(\displaystyle \text{Do I do like this? }\; \frac{\frac{n(n-1)(n-2)(n-3)(n-4)}{5!}} {\frac{n(n-1)(n-2)(n-3)}{4!}} \;=\; \frac{16}{5}\) . Um, not quite

\(\displaystyle \text{If yes, then what next?}\) . How about reducing that fraction?

You forgot about the coefficients in the binomial, 3 and 4.
They affect the answer, too.

\(\displaystyle \text{We have: }\;(4x+3)^n \;=\;{n\choose n}(4x)^n \;+\; {n\choose n-1}(4x)^{n-1}(3) \;+\; {n\choose n-2}(4x)^{n-2}(3^2) \;+\; {n\choose n-3}(4x)^{n-3}(3^3) \;+\; \hdots\)

\(\displaystyle \text{The }x^4\text{-term is: }\;{n\choose4}(4x)^4(3)^{n-4}\)

\(\displaystyle \text{The }x^5\text{-term is: }\;{n\choose5}(4x)^5(3)^{n-5}\)


\(\displaystyle \text{The ratio of their coefficients is: }\;\frac{{n\choose4}\cdot4^4\cdot3^{n-4}}{{n\choose5}\cdot4^5\cdot3^{n-5}} \;=\;\frac{5}{16} \quad\Rightarrow\quad \frac{{n\choose4}\cdot3}{{n\choose5}\cdot 4} \;=\;\frac{5}{16} \quad\Rightarrow\quad 48{n\choose4} \;=\;20{n\choose5}\)

\(\displaystyle \text{We have: }\;48\cdot\frac{n(n-1)(n-2)(n-3)}{4\cdot3\cdot2\cdot1} \;=\;20\cdot\frac{n(n-1)(n-2)(n-3)(n-4)}{5\cdot4\cdot3\cdot2\cdot1}\)


\(\displaystyle \text{Reduce: }\begin{array}{c} \text{In the numerators, cancel }n(n-1)(n-2)(n-3) \\ \text{In the denominators, cancel }4\cdot3\cdot2\cdot1 \end{array}\)

\(\displaystyle \text{And we have: }\;48 \;=\;\frac{20(n-4)}{5} \quad\Rightarrow\quad 12\:=\:n-4 \quad\Rightarrow\quad \boxed{n \:=\:16}\)