Question

How many positive terms are there in the sequence $\left(x_n\right)$ if
$x_n=\frac{195}{4P_n}-\frac{{}^{n+3}{A}_3}{P_{n+1}},n\in N.$

Answer 1

We have,
$x_n=\frac{195}{4P_n}-\frac{{}^{n+3}{A}_3}{P_{n+1}}$
$\therefore x_n=\frac{195}{4.n!}-\frac{(n+3)(n+2)(n+1)}{(n+1)!}$
$=\frac{195}{4.n!}-\frac{(n+3)(n+2)}{n!}$
$=\frac{195-4n^2-20n-24}{4.n!}$
$=\frac{171-4n^2-20n}{4.n!}$
$\because x_n$ is positive.
$\therefore \frac{171-4n^2-20n}{4.n!}>0$
$\Rightarrow 4n^2+20n-171<0$
which is true for $n=1,2,3,4$
Hence the given sequence $\left(x_n\right)$ has $4$ positive terms.