Question

Write the series using summation notation
-2+1+6+13+22+...

Answer 1

`-2+1+6+13+22+...=\sum(n^2+2n-2)`

Explanation:

Series in Summation Notation

We must try to find a general formula for each term and then use summation to generalize the sum of all the terms for any value of n, the term number.

The series is

`-2+1+6+13+22+...`

The difference between consecutive terms will be computed:

`a_2-a_1=1-(-2)=3`

`a_3-a_2=6-1=5`

`a_4-a_3=13-6=7`

We can see that

`a_(n+1)-a_n=2n+1`

Applying summation on both sides:

`\sum [a_(n+1)-a_n]=\sum (2n+1)=\sum 2n+\sum 1`

Knowing that

`\displaystyle \sum n=(n(n+1))/(2)`

`\sum [a_(n+1)-a_n]=n(n+1)+n=n^2+2n`

Similarly, the sum of the left side is found to be

`\sum [a_(n+1)-a_n]=a_(n+1)-a_1`

Replacing into the above equation, we find

`a_(n+1)-a_1=n^2+2n`

Solving

`a_(n+1)=n^2+2n+a_1=n^2+2n-2`

Having the general term, the series is expressed as

`-2+1+6+13+22+...=\sum(n^2+2n-2)`

For n=0 to infinity