a) \(\displaystyle\sum\ {k}={1}\ {n}{k}={n}\ {2}\ {\left({n}\ +\ {1}\right)}\)

b) \(\displaystyle\sum\ {k}={1}\ {n}\ {c}={c}{n}\)

Calculation:

In order to find the sum of an arithmetic sequence, you need

\(\displaystyle{S}\ {n}=\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}\)

Write \(\displaystyle\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}\) as follows,

\(\displaystyle\sum\ {n}={1}\ {100}\ {\left({6}\ -\ {12}{n}\right)}=\sum\ {n}={1}\ {100}\ {6}\ -\ {12}\ \sum\ {n}={1}\ {100}\ {n}\)

Now to find \(\displaystyle\sum\ {n}={1}\ {100}\ {6},\)

Substitute the values of n in equation (b).

We get,

\(\displaystyle\sum\ {n}={1}\ {100}\ {6}\)

\(\displaystyle={6}\ {\left({100}\right)}\)

\(\displaystyle={600}\)

Now to find the \(\displaystyle{12}\ \sum\ {n}={1}\ {100}\ {n}\)

Substitute the values of n in equation (a).

We get,

\(\displaystyle{12}\ \sum\ {n}={1}\ {100}\ {n}\)

\(\displaystyle={12}\ {\left({100}{\left({100}\ +\ {1}\right)}{2}\right)}\)

\(\displaystyle={\left({100}\ \times\ {101}\ {4}\right)}\)

\(\displaystyle={2525}\)

Then,

\(\displaystyle\sum\ {n}={1}\ {100}\ {6}\ -\ {12}\ \sum\ {n}={1}\ {100}\ {n}={600}\ -\ {2525}\)

\(\displaystyle=\ -{1925}\)

Hence

\(\displaystyle\sum\ {n}={1}\ {100}{\left({6}\ -\ {12}{n}\right)}\ =\ -{1925}\)