For positive integer $n$, define

$$ f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} . $$

Then, the value of $$\mathop {\lim }\limits_{n \to \infty } f\left( n \right)$$ is equal to

Let $${f_1}:R \to R,\,{f_2}:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R,\,{f_3}:( - 1,{e^{\pi /2}} - 2) \to R$$ and $${f_4}:R \to R$$ be functions defined by

(i) $${f_1}(x) = \sin (\sqrt {1 - {e^{ - {x^2}}}} )$$,

(ii) $${f_2}(x) = \left\{ \matrix{ {{|\sin x|} \over {\tan { - ^1}x}}if\,x \ne 0,\,where \hfill \cr 1\,if\,x = 0 \hfill \cr} \right.$$

the inverse trigonometric function tan^$$-$$1x assumes values in $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$,

(iii) $${f_3}(x) = [\sin ({\log _e}(x + 2))]$$, where for $$t \in R,\,[t]$$ denotes the greatest integer less than or equal to t,

(iv) $${f_4}(x) = \left\{ \matrix{ {x^2}\sin \left( {{1 \over x}} \right)\,if\,x \ne 0 \hfill \cr 0\,if\,x = 0 \hfill \cr} \right.$$

If f : R $$ \to $$ R is a twice differentiable function such that f"(x) > 0 for all x$$ \in $$R, and $$f\left( {{1 \over 2}} \right) = {1 \over 2}$$, f(1) = 1, then