Answer :
Answer:
[tex]-2+1+6+13+22+...=\sum(n^2+2n-2)[/tex]
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
[tex]-2+1+6+13+22+...[/tex]
The difference between consecutive terms will be computed:
[tex]a_2-a_1=1-(-2)=3[/tex]
[tex]a_3-a_2=6-1=5[/tex]
[tex]a_4-a_3=13-6=7[/tex]
We can see that
[tex]a_{n+1}-a_n=2n+1[/tex]
Applying summation on both sides:
[tex]\sum [a_{n+1}-a_n]=\sum (2n+1)=\sum 2n+\sum 1[/tex]
Knowing that
[tex]\displaystyle \sum n=\frac{n(n+1)}{2}[/tex]
[tex]\sum [a_{n+1}-a_n]=n(n+1)+n=n^2+2n[/tex]
Similarly, the sum of the left side is found to be
[tex]\sum [a_{n+1}-a_n]=a_{n+1}-a_1[/tex]
Replacing into the above equation, we find
[tex]a_{n+1}-a_1=n^2+2n[/tex]
Solving
[tex]a_{n+1}=n^2+2n+a_1=n^2+2n-2[/tex]
Having the general term, the series is expressed as
[tex]-2+1+6+13+22+...=\sum(n^2+2n-2)[/tex]
For n=0 to infinity