Question

Write the sum $2+\frac{2}{4}+\frac{2}{9}+\frac{2}{16}+\frac{2}{25}$in sigma notation in three ways: with a starting value of$\left(a\right)k=1,\left(b\right)k=2,\left(c\right)k=0$

Short answer

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)\sum _{k=1}^5\frac{2}{k^2} \\ \mathrm{Part}\left(b\right)\sum _{k=2}^6\frac{2}{{\left(k-1\right)}^2} \\ \mathrm{Part}\left(c\right)\sum _{k=0}^4\frac{2}{{\left(k+1\right)}^2} \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information 

The given sum is$2+\frac{2}{4}+\frac{2}{9}+\frac{2}{16}+\frac{2}{25}$

Part(a) Step 2. Calculation 

The sum can be written as follows,

$$\begin{gathered} 2+\frac{2}{4}+\frac{2}{9}+\frac{2}{16}+\frac{2}{25}=\frac{2}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{2}{4^2}+\frac{2}{5^2} \\ =\sum _{k=1}^5\frac{2}{k^2} \end{gathered}$$

Part(b) Step 1. Calculation 

The sum can be written as follows,

$$\begin{gathered} 2+\frac{2}{4}+\frac{2}{9}+\frac{2}{16}+\frac{2}{25}=\frac{2}{{\left(2-1\right)}^2}+\frac{2}{{\left(3-1\right)}^2}+\frac{2}{{\left(4-1\right)}^2}+\frac{2}{{\left(5-1\right)}^2}+\frac{2}{{\left(6-1\right)}^2} \\ =\sum _{k=2}^6\frac{2}{{\left(k-1\right)}^2} \end{gathered}$$

Part(c) Step 1. Calculation 

The sum can be written as follows,

$$\begin{gathered} 2+\frac{2}{4}+\frac{2}{9}+\frac{2}{16}+\frac{2}{25}=\frac{2}{{\left(0+1\right)}^2}+\frac{2}{{\left(1+1\right)}^2}+\frac{2}{{\left(2+1\right)}^2}+\frac{2}{{\left(3+1\right)}^2}+\frac{2}{{\left(4+1\right)}^2} \\ =\sum _{k=0}^4\frac{2}{{\left(k+1\right)}^2} \end{gathered}$$