# Finding n given the ratio of nCr to another ratio

• Oct 5th 2009, 05:20 PM
differentiate
Finding n given the ratio of nCr to another ratio
In the expansion of (2+3x)^n the coefficients of x^3 and x^4 are in the raio 8:15. Find n.

$\displaystyle T_3 = \binom{n}{3}2^2(3x)^3$
$\displaystyle T_4 = \binom{n}{4}2^3(3x)^4$

ratios are in 8:15, therefore,
$\displaystyle \frac{108\binom{n}{4}}{648\binom{n}{5}} = \frac{8}{15}$
$\displaystyle \frac{\binom{n}{4}}{6\binom{n}{5}} \frac{8}{15}$
$\displaystyle \frac{\frac{n!}{(n-4)!4!}}{\frac{6n!}{(n-5)!5!}} = \frac{8}{15}$
$\displaystyle \frac{48n!}{(n-5)!5!} = \frac{15n!}{(n-4)!4!}$
$\displaystyle \frac{2n!}{5(n-5)!} = \frac{5n!}{(n-4)!}$
$\displaystyle \frac{2n!}{5(n-5)!} = \frac{5n!}{8(n-4)(n-5)!}$
$\displaystyle \frac{2n!}{5(n-5)!} = \frac{5n!}{8(n-4)(n-5)!}$
$\displaystyle 2n!.8.(n-4) = 5n!.5$
$\displaystyle 16n - 64 = 25$
$\displaystyle 16n= 89$
$\displaystyle n = 5.5625$

BUT OMG the answer is 8!!!!!!!!!!!!!!!!

WHY!! what's wrong with my working out.
• Oct 5th 2009, 07:03 PM
Soroban
Hello, differentiate!

Quote:

In the expansion of $\displaystyle (2+3x)^n$, the coefficients of $\displaystyle x^3$ and $\displaystyle x^4$ are in the raio 8:15.
$\displaystyle \text{Find }n$.

$\displaystyle T_3 = \binom{n}{3}2^2(3x)^3,\quad T_4 = \binom{n}{4}2^3(3x)^4$ . . . . . These are wrong

The two terms are:

. . $\displaystyle T_3 \:=\: {n\choose3}2^{n-3}(3x)^3$

. . $\displaystyle T_4 \:=\: {n\choose4}2^{n-4}(3x)^4$

Then: .$\displaystyle \dfrac{{n\choose3}2^{n-3}(3^3)}{{n\choose4}2^{n-4}(3^4)} \;=\;\frac{8}{15} \quad\Rightarrow\quad \frac{\dfrac{n(n-1)(n-2)}{3!}}{\dfrac{n(n-1)(n-2)(n-3)}{4!}}\cdot\frac{2}{3} \:=\:\frac{8}{15}$

. . . . . . $\displaystyle \frac{4}{n-3}\cdot\frac{2}{3} \:=\:\frac{8}{15} \quad\Rightarrow\quad \frac{1}{n-3} \:=\:\frac{1}{5}$

. . . . . . $\displaystyle 5 \:=\:n-3 \quad\Rightarrow\quad \boxed{n \:=\:8}$