Expert: Paul Klarreich - 4/10/2007

Question
hi i need some clarification with problems involving 'summation' of series.From here on i will use 'E' to represent sigma or summation.this is a problem i did: Given that (r=1 to n)E(r) = (n/2)(n+1), show that (r=1 to n)E(3r+1) = (1/2)n(3n+5).This is the reasoning i used Ur=Sr-S(r-1)where Ur(rth term), Sr and S(r-1)(sum of 1st r terms and sum of 1st (r-1) terms)we get: (1/2)r(3r+5)-[(1/2)(r-1)3(r-1)+5] = (1/2)[3r^2+5r-[(r-1)(3r+2)]]=(1/2)[3r^2-3r^2+5r+r+2]=3r+1 as required(this is true for the second summation (am i correct?) but i did not use the 1st summation as i believe the question required (is this allowed?)). My biggest problem now is how to solve problems such as these (don't know where or how to begin): Express in terms of 'n' and in the simplest form (i)  (r=1 to 2n)E(r) ; (ii)  (r=n+1 to 2n)E(r)  Thank you in advance

Answer
Hi, John,

I am going to change your notation a bit, using SUM instead of your E, and making the dummy variable 'k' instead of 'r'.

So you are starting with these formulas:

WARNING: THIS DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL DIFFICULT TO VIEW ON CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

Well-known formula 1:
n      n(n+1)
SUM k = ------
k=1       2

Well-known formula 2: (Which you didn't write, but I think you can figure it out.)
n
SUM 1 = n
k=1

Now you want to compute this summation.

n
SUM (3k + 1) =
k=1

Use these ideas:

I. SUMMATION of a SUM is the SUM of the SUMMATIONS.

SUM (A + B) = SUM(A) + SUM(B)

II. A constant factor can be factored out.

SUM (cA) = c SUM(A)

Now we can do it:

n
SUM (3k + 1) =
k=1

SUM (3k) + SUM(1) =

3 SUM k + SUM(1) =
 n(n+1)
3 ------ + n
   2

and the rest is just algebra:
3n^2 + 3n
--------- + n =
   2
3n^2 + 3n + 2n
-------------- =
   2
3n^2 + 5n
---------- =
   2
n(3n + 5)
----------
   2