Question

For positive integer $n$, define $f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-n^2}{12n+3n^2}+...+\frac{25n-7n^2}{7n^2}$
Then, the value of $\underset{n\to \infty }{lim}f\left(n\right)$ is eequal to

• A. $4-\frac{3}{4}{\log }_e\left(\frac{7}{3}\right)$
• B. $4-\frac{4}{3}{\log }_e\left(\frac{7}{3}\right)$
• C. $3+\frac{3}{4}{\log }_{e}7$
• D. $3+\frac{4}{3}{\log }_{e}7$

Answer 1

The correct option is A $4-\frac{3}{4}{\log }_e\left(\frac{7}{3}\right)$

$f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\cdots +\frac{25n-7n^2}{7n^2}$

$=(\frac{16+5n-3n^2}{4n+3n^2}+1)+(\frac{32+n-3n^2}{8n+3n^2}+1)+\cdots +(\frac{25n-7n^2}{7n^2}+1)$


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

$=\sum _{r=1}^n\frac{9n+16r}{4rn+3n^2}=\frac{1}{n}\sum _{r=1}^n\frac{9+16\left(\frac{r}{n}\right)}{4\left(\frac{r}{n}\right)+3}$

$\underset{n\to \infty }{lim}f\left(n\right)=\underset{0}{\overset{1}{\int }}\frac{9+16x}{4x+3}dx$

$=\underset{0}{\overset{1}{\int }}\frac{(16x+12)-3}{4x+3}dx$

$={[4x-\frac{3}{4}\ln |4x+3|]}_0^1$

$=4-\frac{3}{4}\ln \frac{7}{3}$