1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sequences and Series

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the sum of the products of every pair of the first 'n' natural numbers.

    2. Relevant equations
    sigma n^2 = n(n+1)(2n+1)/6

    3. The attempt at a solution
    S=1.2 + 1.3 + 1.4 .....+ 2.3 + 2.4 ........n-1(n)
    i cant figure out how to proceed ..
  2. jcsd
  3. Nov 24, 2011 #2


    User Avatar
    Science Advisor
    Homework Helper

    Think about expanding the expression (1+2+...+n)*(1+2+...+n) like it was a polynomial.
  4. Nov 24, 2011 #3
    (1+2+...n)^2 = (1.1 +2.2 ......n.n) + (1.2+1.3 ...1.n .....n-1.n)
    please guide me further
  5. Nov 24, 2011 #4


    User Avatar
    Science Advisor
    Homework Helper

    You have a formula for summing the squares 1^2+2^2+...+n^2. You've probably got a formula for summing 1+2+..+n as well. Use it.
  6. Nov 24, 2011 #5


    User Avatar
    Gold Member

    That's a good question and very conceptual. Try to break your problem into a general summation which can be expressed by a variable.


    Don't you worry about n(0), that thing is created to be zero, as you'll see in next step.

    We can take the numbers outside the brackets as r, whose value varies from r=1,2,...,(n-1),n.

    Now about the summation of numbers inside the bracket. You can take out their sum by AP. Since the first term of any bracket is (r+1) and the last term is always n, the number of terms can be figured as (n-r), since you're not counting first r natural number.

    Therefore, our general term can be written as:


    NOTE: Try considering r=n. It is a huge relief that Tn=0, or else we would have to manually subtract it in the end. Always consider manually checking last term in the general term, they may give you a problem.

    Now fearlessly take summation of our series.


    Now open it and you'll get the following:


    Simple equation in summation of r, r2 and r3
  7. Nov 25, 2011 #6
    Thanx for the reply
    summing up your equation , i got
    n[n+1/2]^2 [ (3n-2)(n-1) / 12 ]
    will you check if this right . . . .
  8. Nov 25, 2011 #7
    i am told that this is a better way to do it..
    but i cant begin to undrstand the (sigma n)^2 - sigma n^2 thing

    Attached Files:

    • Sol.PNG
      File size:
      38.5 KB
  9. Nov 25, 2011 #8


    User Avatar
    Gold Member


    This is a better solution, but I was afraid of it since I was not fully sure whether it will work or not, but I guess in the end, my hunch was right! :biggrin:

    Its simple enough. Let's see.




    As you can see, after multiplying, some squares are obtained, which we remove by Ʃr2.

    Further more, You can see every non-square term will repeat two time, so we divide the remaining part by 2 and here we are.
  10. Nov 25, 2011 #9
    Thanx AGNuke , that was some real help !
  11. Nov 26, 2011 #10


    User Avatar
    Gold Member

    Welcome! :wink:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Sequences and Series
  1. Series and sequence (Replies: 3)

  2. Sequences and Series (Replies: 1)

  3. Sequences and Series (Replies: 1)

  4. Sequence and series (Replies: 1)

  5. Series sequences (Replies: 9)