Question

How do you evaluate `\sum _ { k = 1} ^ { 6} ( 4k - 4)`?

Answer 1

I got `60`

Explanation:

The terms of your Arithmetic Progression are:
`\sum_{k=1}^6(4k-4)=(4*1-4)+(4*2-4)+(4*3-4)+(4*4-4)+(4*5-4)+(4*6-4)=`
`=0+4+8+12+16+20`

Arithmetic Progression of `n=6` terms with ratio `r=4`.

The sum will be:

`S=n/2(a_1+a_2)`

Where:
`a_1=0` is the first term;
`a_n=a_6=20` is the last term.

We get: `S=6/2(0+20)=3*20=60`