Question 310266
Please help me solve this...
write the following in summation notation
1+6+13+22...
<pre><font size = 4 color = "indigo"><b>
I assume you want it in the form {{{sum(a[n], n=1, infinity)}}}

First we find the general term by assuming its general term
can be expressed as a polynomial.  To find the smallest
possible degree it can have we write the four given numbers in
a line like this

1   6   13   22 

Then we find a line of differences by subtracting each number 
from the one just to the right of it and writing the difference 
between and below the numbers subtracted, like this:

1   6   13   22
  5   7    9

5, 7, and 9 are not all the same number, so we do the same to 
the bottom line:

1   6   13   22
  5   7    9
    2    2

Those are the same, since they are both 2's.  Since it requires 2 lines
of differences to get them all to be the same number, we assume the general
term is a polynomial of degree 2:

{{{a[n]=An^2+Bn+C}}}

We substitute n=1,2, and 3

{{{system(a[1]=A(1)^2+B(1)+C,
a[2]=A(2)^2+B(2)+C,
a[3]=A(3)^2+B(3)+C)}}}  

{{{system(1=A+B+C,
6=4A+2B+C,
13=9A+3B3+C)}}}

{{{system(A+B+C=1,
4A+2B+C=6,
9A+3B3+C=13)}}}

We solve that system and get

{{{A=1}}}, {{{B=2}}}, and {{{C=-2}}}

So,

{{{a[n]=1n^2+2n-2}}}

{{{a[n]=n^2+2n-2}}}

So we write:

{{{sum(a[n], n=1, infinity) = sum( (n^2+2n-2), n=1, infinity )}}}

Note: The series could also be written in the form:

{{{sum(a[n], n=0, infinity) = sum( (n^2+4n+1), n=0, infinity )}}}

where the index (or "dummy variable") n starts at 0.  Then the
system of equations would be easier to solve.  But I am guessing 
that your teacher probably wanted the summation index to start at
1, not 0.

Edwin</pre>