Question

Verify that$\underset{k=1}{\overset{n}{\sum k}}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{n(n+1)}{2}$for the cases$\left(a\right)n=2,\left(b\right)n=8,\left(c\right)n=9$

Short answer

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)Forn=2,\underset{k=1}{\overset{n}{\sum k}}=\frac{n(n+1)}{2} \\ \mathrm{Part}\left(b\right)Forn=8,\underset{k=1}{\overset{n}{\sum k}}=\frac{n(n+1)}{2} \\ \mathrm{Part}\left(c\right)Forn=9,\underset{k=1}{\overset{n}{\sum k}}=\frac{n(n+1)}{2} \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information  

The given sum is$\underset{k=1}{\overset{n}{\sum k}}$

Part(a) Step 2. Calculation  

The left-hand side sum and the right-hand side sum is as follows,

$$\begin{gathered} \underset{k=1}{\overset{2}{\sum k}}=1+2 \\ =3 \\ and \\ \frac{2(2+1)}{2}=\frac{2\left(3\right)}{2} \\ =3 \end{gathered}$$

Thus, it is verified.

Part(b) Step 1. Calculation  

The left-hand side sum and the right-hand side sum are as follows,

$$\begin{gathered} \underset{k=1}{\overset{8}{\sum k}}=1+2+3+4+5+6+7+8 \\ =36 \\ and \\ \frac{8(8+1)}{2}=\frac{8\left(9\right)}{2} \\ =36 \end{gathered}$$

Thus, it is verified.

Part(c) Step 1. Calculation  

The left-hand side sum and the right-hand side sum are as follows,

$$\begin{gathered} \underset{k=1}{\overset{9}{\sum k}}=1+2+3+4+5+6+7+8+9 \\ =45 \\ and \\ \frac{9(9+1)}{2}=\frac{9\left(10\right)}{2} \\ =45 \end{gathered}$$

Thus, it is verified.