Question

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

Answer 1

Answer:

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

Step-by-step 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

$\displaystyle \sum n=\frac{n(n+1)}{2}$

$\sum [a_{n+1}-a_n]=n(n+1)+n=n^2+2n$

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