Question

Find the number of negative terms of the sequence
$x_n=\frac{{}^{n+4}{P}_4}{P_{n+2}}-\frac{143}{4P_n}$

Answer 1

We have,
$x_n=\frac{{}^{n+4}{P}_4}{P_{n+2}}-\frac{143}{4P_n}$
$\therefore x_n=\frac{(x+4)(n+3)(n+2)(n+1)}{(n+2)!}-\frac{143}{4.n!}$
$=\frac{(x+4)(n+3)(n+2)(n+1)}{(n+2)(n+1)n!}-\frac{143}{4.n!}$
$=\frac{(x+4)(n+3)}{n!}-\frac{143}{4.n!}$
$=\frac{(4n^2+28n-95)}{4.n!}$
$\because x_n$ is negative.
$\therefore =\frac{(4n^2+28n-95)}{4.n!}<0$
$\Rightarrow =(4n^2+28n-95)<0$
which is true for $n=1,2$
Hence, $x_1=-\frac{63}{4}$ and $x_2=-\frac{23}{8}$ are two negative terms.