Question

(1!/n!)-(3/(n+1)!)-(n^2-4/(n+2)!)

Answer 1

Answer:

BASIC FORMULA USED :-)

$(n+1)! =( n+1)n!$

Now using above formula,

Value of given expression can be calculated as:-

$\frac{1 !}{n!} - \frac{3}{(n + 1)!} - \frac{ {n}^{2} - 4}{(n + 2)!} \\ \\ = \frac{1}{n!} - \frac{3}{(n + 1)n ! } - \frac{ {n}^{2} - 4}{(n + 2)(n + 1)n!} \\ \\ = \frac{ (n + 1)(n + 2) - 3(n + 2) - (n {}^{2} - 4)}{(n + 2)(n + 1)n ! } \\ \\ = \frac{ n {}^{2} + 3 n + 2 - 3n - 6 - n {}^{2} + 4}{(n + 2) ! } \\ \\ = \frac{0}{(n + 2) ! } \\ \\ = 0$

So,

The final value of given expression is zero .