Question

In the expansion of $(1+x{)}^n$ by the increasing powers of $x$, the third term is four times as great as the fifth term, and the ratio of the fourth term to the sixth is $\frac{40}{3}$. Find $n$ and $x$.

Answer 1

$(1+x{)}^n$

3rd term${=}^n{C}_2{x}^{n-3}$

5th term${=}^n{C}_4{x}^{n-5}$

${}^n{C}_2{x}^{n-3}=4^n{C}_4{x}^{n-5}$

$\Rightarrow \frac{n!}{(n-2)!2!}{x}^2=\frac{4n!}{(x-4)!4!}$

$\Rightarrow \frac{x^2}{(n-2)(n-3)(n-4)!2!}=\frac{4}{(x-4)!4\times 3\times 2!}$

$\Rightarrow 3x^2=(n-2)(n-3)$

$\Rightarrow 3x^2=n^2-5n+6$---------------(1)

$\frac{4thterm}{6thterm}=\frac{40}{3}$

$\Rightarrow \frac{{}^n{C}_3{x}^{n-4}}{{}^n{C}_5{x}^{n-6}}=\frac{40}{3}$

$\Rightarrow \frac{5\times 4\times 3!\times (n-5)!x^2}{(n-3)(n-4)(n-5)!3!}=\frac{40}{3}$

$\Rightarrow 3x^2=2(n-3)(n-4)$

$\Rightarrow 3x^2=2(n-3)(n-4)$

$\Rightarrow 3x^2=2n^2-14n+24$-------------(2)

From Equation 1& 2,

$n^2-5n+6=2n^2-14n+24$

$\Rightarrow n^2-9n+18=0$

$\Rightarrow n^2-6n-3n+18=0$

$\Rightarrow (n-6)(n-3)=0$

$\Rightarrow n=3$ or $n=6$.

$n\ne 3$, then $x=0$

From, Equation 1

$3x^2=n^2-5n+6$

$\Rightarrow 3x^2=6^2-6\cdot 5+6$o,

$\Rightarrow 3x^2=36-30+6$

$\Rightarrow 3x^2=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

So, $n=3,x=0$

and $n=6,x=\pm 2$