Finding n given the ratio of nCr to another ratio

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October 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.

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

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

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

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

Hello, differentiate!

Quote:

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

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

The two terms are:

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

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

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

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

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

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