Question

Show that$\sum _{k=0}^8\frac{1}{k^2+1}\mathrm{is}\mathrm{equal}\mathrm{to}2\sum _{k=0}^8\frac{1}{2k^2+2}$by writing out the terms in each sum.

Short answer

It is shown that$\sum _{k=0}^8\frac{1}{k^2+1}=2\sum _{k=0}^8\frac{1}{2k^2+2}$

Step-by-step solution

Step 1. Given Information    

The given summations are $\sum _{k=0}^8\frac{1}{k^2+1}and2\sum _{k=0}^8\frac{1}{2k^2+2}$

Step 2. Explanation  

The first sum can be written as follows,

$$\begin{gathered} \sum _{k=0}^8\frac{1}{k^2+1}=1+\frac{1}{1+1}+\frac{1}{4+1}+\frac{1}{9+1}+\frac{1}{16+1}+\frac{1}{25+1}+\frac{1}{36+1}+\frac{1}{49+1}+\frac{1}{64+1} \\ =1+\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\frac{1}{26}+\frac{1}{37}+\frac{1}{50}+\frac{1}{65} \end{gathered}$$

and the second sum can be written as,

$$\begin{gathered} 2\sum _{k=0}^8\frac{1}{2k^2+2}=2\left(\frac{1}{0+2}+\frac{1}{2{\left(1\right)}^2+2}+\frac{1}{2{\left(2\right)}^2+2}+\frac{1}{2{\left(3\right)}^2+2}+\frac{1}{2{\left(4\right)}^2+2}+\frac{1}{2{\left(5\right)}^2+2}+\frac{1}{2{\left(6\right)}^2+2}+\frac{1}{2{\left(7\right)}^2+2}+\frac{1}{2{\left(8\right)}^2+2}\right) \\ =2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{10}+\frac{1}{20}+\frac{1}{34}+\frac{1}{52}+\frac{1}{74}+\frac{1}{100}+\frac{1}{130}\right) \\ =1+\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\frac{1}{26}+\frac{1}{37}+\frac{1}{50}+\frac{1}{65} \end{gathered}$$

Hence, we can say that both the summations are equal