Section 4.2 #24: Compute $$\displaystyle\sum_{k=1}^n \dfrac{2k^3-3k}{n^2}$$ for $n=10$, $100$, $1000$, and $10000$.
Solution: Using the properties of summations and the formulas in Theorem 4.2: $$\begin{array}{ll} \displaystyle\sum_{k=1}^n \dfrac{2k^3 - 3k}{n^2} &= \dfrac{2}{n^2} \left( \displaystyle\sum_{k=1}^n k^3 \right) - \dfrac{3}{n^2} \left( \displaystyle\sum_{k=1}^n k \right) \\ &= \dfrac{2}{n^2} \left( \dfrac{n^2(n+1)^2}{4} \right) - \dfrac{3}{n^2} \left( \dfrac{n(n+1)}{2} \right) \\ \end{array}$$ Therefore we see for $n=10$, the sum equals $$\dfrac{2}{100} \left( \dfrac{100(121)}{4} \right) - \dfrac{3}{100} \left( \dfrac{10(11)}{2} \right)= \dfrac{121}{2} - \dfrac{33}{20} = \dfrac{1177}{20}.$$ For $n=100$, the sum equals $\dfrac{1019797}{2}$, for $n=1000$ the sum equals $5009989985$, and for $n=10000$ the sum equals $50009998999850$.
Section 4.2 #42: Compute $$\displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n \left( 2 + \dfrac{3k}{n} \right)^3 \dfrac{3}{n}.$$
Solution: First expand the binomial cubed to get $$\left( 2 + \dfrac{3k}{n} \right)^3 = \dfrac{27k^3}{n^3} + \dfrac{54k^2}{n^2} + \dfrac{36k}{n} +8.$$ Therefore using the properties of sums and Theorem 4.2, we get $$\begin{array}{ll} \displaystyle\sum_{k=1}^n \left( 2+\dfrac{3k}{n} \right)^3 \dfrac{3}{n} &= \dfrac{3}{n} \left[ \dfrac{27}{n^3} \left( \displaystyle\sum_{k=1}^n k^3 \right) + \dfrac{54}{n^2} \left( \displaystyle\sum_{k=1}^n k^2 \right) + \dfrac{36}{n} \left( \displaystyle\sum_{k=1}^n k \right) + 8 \left( \displaystyle\sum_{k=1}^n 1 \right) \right] \\ &= \dfrac{3}{n} \left[ \dfrac{27}{n^3} \left( \dfrac{n^2(n+1)^2}{4} \right) + \dfrac{54}{n^2} \left( \dfrac{n(n+1)(2n+1)}{6} \right) + \dfrac{36}{n} \left( \dfrac{n(n+1)}{2} \right) + 8n \right] \\ &= \dfrac{81}{4} \dfrac{n^2+2n+1}{n^2} + 27 \dfrac{2n^2+3n+1}{n^2} + 54 \dfrac{n+1}{n} + 24 \\ &= \dfrac{81}{4} \left( 1 + \dfrac{2}{n} + \dfrac{1}{n^2} \right) + 27 \left( 2 + \dfrac{3}{n} + \dfrac{1}{n^2} \right) + 54 \left( 1 + \dfrac{1}{n} \right) + 24. \end{array}$$ Therefore $$\begin{array}{ll} \displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n \left( 2 + \dfrac{3k}{n} \right)^3 \dfrac{3}{n} &= \displaystyle\lim_{n \rightarrow \infty} \dfrac{81}{4} \left( 1 + \dfrac{2}{n} + \dfrac{1}{n^2} \right) + 27 \left( 2 + \dfrac{3}{n} + \dfrac{1}{n^2} \right) + 54 \left( 1 + \dfrac{1}{n} \right) + 24 \\ &= \dfrac{81}{4} + 2(27) + 54 + 24 \\ &= \dfrac{609}{4}. \end{array}$$