Question

$\displaystyle \rm\lim_{n \to \infty } \sum_{i = 1}^{n} \frac{ \left(2 + \frac{i}{n} \right)^{2} + \left(2 + \frac{ i }{n} \right) }{n}$​

Answer 1

Answer:

The answer is $\dfrac{53}{6}$

Step-by-step explanation:

    $\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{(2+\frac{i}{n})^2+(2+\frac{i}{n})}{n}$

   $=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{4+\frac{4i}{n}+\frac{i^2}{n^2}+(2+\frac{i}{n})}{n}\\\\=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{6+\frac{5i}{n}+\frac{i^2}{n^2}}{n}\\\\=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{6n^2+5ni+i^2}{n^3}\\\\=\displaystyle \lim_{n\to \infty}\dfrac{1}{n^3} (6n^2\sum_{i=1}^n1+5n\sum_{i=1}^ni+\sum_{i=1}^ni^2)\\\\=\displaystyle \lim_{n\to \infty}\dfrac{1}{n^3} (6n^2(n)+5n\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6})$

   $=\displaystyle \lim_{n\to \infty}(6+\frac{5}{2}\frac{(n+1)}{n}+\frac{1}{6}\frac{(n+1)(2n+1)}{n^2})\\\\=\displaystyle \lim_{n\to \infty}(6+\frac{5}{2}(1+\frac{1}{n})+\frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n}))\\\\\\=6+\dfrac{5}{2}+\dfrac{2}{6}=\dfrac{53}{6}$