Mathematics - Permutation & Combination Question with Solution | TestHub

We have, $x_n=\frac{195}{4{}^n{P}_n}-\frac{{}^{n+3}{P}_3}{{}^{n+1}{P}_{n+1}},n\in N$

$\begin{array}{rlr}& \frac{195}{4.n!}-\frac{(n+3)(n+2)(n+1)}{(n+1)!}=& \frac{171-4n^2-20n}{4.n!}\end{array}$ $\therefore x^{x_n}$ is positive $\therefore \frac{171-4n^2-20n}{4n!}>0\Rightarrow 4n^2+20n-171<0$

Which is true for $n=1,2,3,4$.

Hence, the given sequence has 4 positive terms.