Tính 

D

- Hướng dẫn giải

Đáp án D

Ta có

$\begin{array}{l}\frac{2k^3+8k^2+6k-1}{k^2+4k+3}\\ =\frac{2k\left(k+1\right)\left(k+3\right)}{\left(k+1\right)\left(k+3\right)}-\frac{1}{\left(k+1\right)\left(k+3\right)}\\ =2k-\frac{1}{2}\left(\frac{1}{k+1}-\frac{1}{k+3}\right)\end{array}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^n\frac{2k^3+8k^2+6k-1}{k^2+4k+3}\\ =\sum _{k=1}^n\left[2k-\frac{1}{2}\left(\frac{1}{k+1}-\frac{1}{k+3}\right)\right]\end{array}$

$=2.\left(1+2+\mathrm{...}+n\right)-\frac{1}{2}\left[\left(\frac{1}{1+1}-\frac{1}{1+3}+\mathrm{...}+\frac{1}{n-1+1}-\frac{1}{n-1+3}+\frac{1}{n+1}-\frac{1}{n+3}\right)\right]$

$$\begin{gathered} =2\frac{n\left(n+1\right)}{2}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3}\right) \\ =n\left(n+1\right)-\frac{1}{2}\left(\frac{5}{6}-\frac{2n+5}{\left(n+2\right)\left(n+3\right)}\right) \end{gathered}$$

suy ra 

$\begin{array}{l}\lim \left(n^2+n-\sum _{k=1}^n\frac{2k^3+8k^2+6k-1}{k^2+4k+3}\right)\\ =\lim \left[\left(n^2+n\right)-\left(n^2+n-\frac{5}{12}+\frac{2n+5}{2\left(n+2\right)\left(n+3\right)}\right)\right]\end{array}$

$\begin{array}{l}=\lim \left(\frac{5}{12}-\frac{2n+5}{2n^2+10n+12}\right)\\ =\lim \frac{5}{12}-\lim \frac{\frac{2}{n}+\frac{5}{n^2}}{2+\frac{10}{n}+\frac{12}{n^2}}=\frac{5}{12}\end{array}$