Binomial Theorem - Relationship between coefficients question (1 Viewer)

Casson8 · Dec 15, 2011

I'm stumped on this question and how you would explain the steps in obtaining the end answer:

"The coefficients of x⁴ and x⁵ are equal in the expansion of (1+x)ⁿ. Evaluate n."

Thanks in advance.

nightweaver066 · Dec 15, 2011

$T_{n+1} = ^nC_rx^{r}$

For term with x^5, r = 5

$T_{n+1} = ^nC_5x^{5}$

For term with x^4, r = 4

$T_{n+1} = ^nC_4x^{4}$

As the coefficients are equal, $^nC_4 = ^nC_5$

$\frac{n!}{(n-4)!} = \frac{n!}{(n-5)!}$

$\frac{1}{(n-4)!} = \frac{1}{(n-5)!}$

$n-5 = 1$

$n = 6$

Shadowdude · Dec 15, 2011

nightweaver066 said:
$T_{n+1} = ^nC_rx^{r}$

For term with x^5, r = 5

$T_{n+1} = ^nC_5x^{5}$

For term with x^4, r = 4

$T_{n+1} = ^nC_4x^{4}$

As the coefficients are equal, $^nC_4 = ^nC_5$

$\frac{n!}{(n-4)!} = \frac{n!}{(n-5)!}$

$\frac{1}{(n-4)!} = \frac{1}{(n-5)!}$

$n-5 = 1$

$n = 6$

$\frac{n!}{(n-k)!}$ is the formula for P(n,k) not C(n,k)

From guesswork, the answer you should need is n = 9, because C(n,4) = C(n,5) in that case.

Trebla · Dec 15, 2011

nightweaver066 said:
$T_{n+1} = ^nC_rx^{r}$

For term with x^5, r = 5

$T_{n+1} = ^nC_5x^{5}$

For term with x^4, r = 4

$T_{n+1} = ^nC_4x^{4}$

As the coefficients are equal, $^nC_4 = ^nC_5$

$\frac{n!}{4!(n-4)!} = \frac{n!}{5!(n-5)!}$

$\frac{5}{(n-4)!} = \frac{1}{(n-5)!}$

$n-4 = 5$

$n = 9$

Corrected

Binomial Theorem - Relationship between coefficients question (1 Viewer)

Casson8

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Shadowdude

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