For positive integer n , define f(n) = n + 16 + 5n - 3 n ^ 2 4n + 3...

Given $f\left(n\right) = n + \dfrac{16 + 5n - 3{n}^{2}}{4n + 3{n}^{2}} + \dfrac{32 + n - 3{n}^{2}}{8n + 3{n}^{2}} + \ldots . + \dfrac{25n - 7{n}^{2}}{7{n}^{2}}$

$= \left(\dfrac{16 + 5n - 3{n}^{2}}{4n + 3{n}^{2}} + 1\right) + \left(\dfrac{32 + n - 3{n}^{2}}{8n + 3{n}^{2}} + 1\right) + \ldots \ldots + \left(\dfrac{25n - 7{n}^{2}}{7{n}^{2}} + 1\right)$

$f\left(n\right) = \dfrac{9n + 16}{4n + 3{n}^{2}} + \dfrac{9n + 32}{8n + 3{n}^{2}} + \ldots \ldots + \dfrac{25n}{7{n}^{2}}$

i.e. $f\left(n\right) = \mathop{∑}_{r = 1}^{n}\dfrac{9n + 16r}{4rn + 3{n}^{2}} = \dfrac{1}{n}\mathop{∑}_{r = 1}^{n}\dfrac{9 + 16\left(\dfrac{r}{n}\right)}{4\left(\dfrac{r}{n}\right) + 3}$

Now, $\lim_{n \to \infty}f\left(n\right) = {∫}_{0}^{1}\dfrac{9 + 16x}{4x + 3}dx$

$= {∫}_{0}^{1}\dfrac{\left(16x + 12\right) - 3}{4x + 3}dx$

= {\left\left(4x - \dfrac{3}{4}\ln \left|4x + 3\right|\right)\right|}_{0}^{1} $= 4 - \dfrac{3}{4}ln\dfrac{7}{3}$