Chapter 11, Problem 94RE

a.

To determine

To rewrite: The sum as a rational function $S(n)$ .

Answer to Problem 94RE

  $S(n)=\frac{-3(n+1)(n+9)}{4n^2}$

Explanation of Solution

Given Information: Given sum is ${\displaystyle \sum _{i=1}^n\left[4-{\left(\frac{3i}{n}\right)}^2\right]{\left(\frac{3i}{n}\right)}^2}$ .

Calculation:

  $S(n)={\displaystyle \sum _{i=1}^n\left[4-{\left(\frac{3i}{n}\right)}^2\right]\left(\frac{3i}{n^2}\right)}$

  $=4\left(\frac{3}{n^2}\right){\displaystyle \sum _{i=1}^{n}i}-\left(\frac{3^3}{n^4}\right){\displaystyle \sum _{i=1}^n{i}^3}$

  $=\frac{12}{n^2}\frac{n(n+1)}{2}-\frac{27}{n^4}{\left(\frac{n(n+1)}{2}\right)}^2$

  $=\frac{6(n+1)}{n}-\frac{27{(n+1)}^2}{4n^2}$

  $=\frac{3(n+1)}{n}\left(2-\frac{9(n+1)}{4n}\right)$

  $=\frac{3(n+1)}{n}\left(\frac{8n-9n-9}{4n}\right)$

  $=\frac{3(n+1)}{n}\frac{(-n-9)}{4n}$

  $=\frac{-3(n+1)(n+9)}{4n^2}$

Thus, $S(n)=\frac{-3(n+1)(n+9)}{4n^2}$ .

b.

To determine

To complete: The given table using $S(n)$ .

Answer to Problem 94RE

  $\begin{array}{llllll}n\hfill & {10}^0\hfill & {10}^1\hfill & {10}^2\hfill & {10}^3\hfill & {10}^4\hfill \\ S(n)\hfill & -15\hfill & -1.5675\hfill & -0.825675\hfill & -0.7575\hfill & -0.75075\hfill \end{array}$

Explanation of Solution

Given Information: An incomplete table is given.

Table:

  $\begin{array}{llllll}n\hfill & {10}^0\hfill & {10}^1\hfill & {10}^2\hfill & {10}^3\hfill & {10}^4\hfill \\ S(n)\hfill & -15\hfill & -1.5675\hfill & -0.825675\hfill & -0.7575\hfill & -0.75075\hfill \end{array}$

c.

To determine

To find: $\underset{n\to \infty }{\lim }S(n)$.

Answer to Problem 94RE

  $\underset{n\to \infty }{\lim }S(n)=-0.75$

Explanation of Solution

Given Information: ${\displaystyle \sum _{i=1}^n\left[4-{\left(\frac{3i}{n}\right)}^2\right]{\left(\frac{3i}{n}\right)}^2}$ and $S(n)$ from part (a).

Calculation:

  $\begin{array}{l}\underset{n\to \infty }{\lim }S(n)=\underset{n\to \infty }{\lim }\frac{-3(n+1)(n+9)}{4n^2}\\ \\ =\underset{n\to \infty }{\lim }\frac{-3}{4}\left(1+\frac{1}{n}\right)\left(1+\frac{9}{n}\right)\\ \\ =\frac{-3}{4}\end{array}$

Thus, $\underset{n\to \infty }{\lim }S(n)=-0.75$ .