Expert: Steve Holleran - 8/16/2008

Question
State the formulae for:
n E r^2 +  n E r      
 n=1       n=1
(The n should be above the "E")
Use these formulae to find and expression for the sum to n of these series.
(1x2)+ (3x4) +(5x6) + (7x8).....
Give your answer in fully factorized form.

I understand everything up to the point of slotting in the terms n(n+1)(2n+1)  +  n(n+1)
     ------------     ------
          6              2   
But i think i am doing something wrong from this line onwards when i say: n  (n+1)+(2n+1) + n(n+1)
                   -
                   3
I think its probably the denominators. Please could you help? Thank you.

Answer
Hi Aisha,

Well, for the first part of the problem that you mention, if you are trying to add the results for the sum of the first "n" squares and the first "n" integers, you have

 [n(n+1)(2n+1)]/6  +  [n(n+1)/2]

= [n(n+1)(2n+1)]/6  + [3n(n+1)/6]    (Mult top and bottom of second
                                     bracket by 3)

= [n(2n^2 + 3n + 1)/6] + [(3n^2 + 3n)/6]

= [(2n^3 + 3n^2 + n)/6] + [(3n^2 + 3n)/6]

= [2n^3 + 6n^2 + 4n] / 6 = [2n(n+2)(n+1)]/6 = [n(n+1)(n+2)/3]

This would be the factored form of the result.

Now for the bad news:  I have no idea how you are supposed to use this result to find the sum of the first n terms of the other series you mention.

I can see how this series can be written as E(k=1 to n) [(2k-1)* x^(2k)], but I don't see how you can sum this using the formulas up top.

Maybe I'm missing something about the exercise.

Steve