Question

Write the sum $\frac{4}{7}+\frac{5}{8}+\frac{6}{9}+\frac{7}{10}+\frac{8}{11}$in sigma notation in three ways: with a starting value of$\left(a\right)k=4,\left(b\right)k=7,\left(c\right)k=5$

Short answer

$$\begin{gathered} \mathrm{Part}\left(a\right)\sum _{k=4}^8\frac{k}{k+3} \\ \mathrm{Part}\left(b\right)\sum _{k=7}^{11}\frac{k-3}{k} \\ \mathrm{Part}\left(c\right)\sum _{k=5}^9\frac{k-1}{k+2} \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information 

The given sum is$\frac{4}{7}+\frac{5}{8}+\frac{6}{9}+\frac{7}{10}+\frac{8}{11}$

Part(a) Step 2. Calculation

The sum can be written as follows,

$$\begin{gathered} \frac{4}{7}+\frac{5}{8}+\frac{6}{9}+\frac{7}{10}+\frac{8}{11}=\frac{4}{4+3}+\frac{5}{5+3}+\frac{6}{6+3}+\frac{7}{7+3}+\frac{8}{8+3} \\ =\sum _{k=4}^8\frac{k}{k+3} \end{gathered}$$

Part(b) Step 1. Calculation

The sum can be written as follows,

$$\begin{gathered} \frac{4}{7}+\frac{5}{8}+\frac{6}{9}+\frac{7}{10}+\frac{8}{11}=\frac{7-3}{7}+\frac{8-3}{8}+\frac{9-3}{9}+\frac{10-3}{10}+\frac{11-3}{11} \\ =\sum _{k=7}^{11}\frac{k-3}{k} \end{gathered}$$

Part(c) Step 1. Calculation

The sum can be written as follows,

$$\begin{gathered} \frac{4}{7}+\frac{5}{8}+\frac{6}{9}+\frac{7}{10}+\frac{8}{11}=\frac{5-1}{5+2}+\frac{6-1}{6+2}+\frac{7-1}{7+2}+\frac{8-1}{8+2}+\frac{9-1}{9+2} \\ =\sum _{k=5}^9\frac{k-1}{k+2} \end{gathered}$$