How Can You Derive the Formula 0.5(n+1)(n+2) from a Summation Problem?

  • Thread starter Thread starter jorgen
  • Start date Start date
  • Tags Tags
    Sum
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 4K views
jorgen
Messages
12
Reaction score
0
Hi all,

I have the following sum

[tex]\sum n - n_1 + 1[/tex]

which I split up in two independent sums

[tex]\sum_{n_1=0}^N n + 1 - \sum_{n_1=0}^N n_1[/tex]

the last sum can be written as

[tex]0.5*n(n+1)[/tex]

but how to rewrite the first sum any hints appreciated. The final answer is
0.5(n+1)(n+2)

but as stated above I have some problems getting there.
Any help or advice appreciated thanks in advance
 
Physics news on Phys.org
Unless you are holding something back,
[tex]\sum_{n_1=0}^N n + 1 = (N + 1)(n + 1)[/tex]

Further,
[tex]\sum_{n_1=0}^N n_1 = \frac12 N (N + 1)[/tex]
(note the capital N); assuming that in the first line,
[tex]\sum n - n_1 + 1[/tex]
you meant
[tex]\sum_{n_1 = 1}^N n - n_1 + 1[/tex]
you are otherwise more or less correct...

Your notation is confusing though.
 
thanks for the reply - my problem is understanding the first summation

[tex](N+1)(n+1)[/tex]

how is small n to be interpreted?

Thanks in advance any hints appreciated.
 
jorgen said:
thanks for the reply - my problem is understanding the first summation

[tex](N+1)(n+1)[/tex]

how is small n to be interpreted?

Thanks in advance any hints appreciated.
However YOU mean it! You wrote
[tex]\sum_{n1= 0}^N (n+1)[/tex]
The "index" is n1 and that changes from 0 to N, but there is no "n1" in the sum itself- you are just adding the number n+ 1 to itself N+1 times. Any number added to itself N+1 times is just N+1 times that number: here (N+1)(n+1).

Actually it seems peculiar to me to use "n1" as an index. Why the 1? You are, of course, welcome to use whatever labels you like but I would have thought that [itex]\sum n- n1+ 1[/itex] would be interpreted as
[tex]\sum_{n=0}^N n- n1+ 1[/itex] <br /> where n1 is some fixed number.[/tex]
 
thanks,

n_1 can change its value that is why I write it like that. So in order to get to

0.5(n+1)(n+2)

I have to say that in the limit n = N?

thanks in advance
 
jorgen said:
thanks,

n_1 can change its value that is why I write it like that. So in order to get to

0.5(n+1)(n+2)

I have to say that in the limit n = N?

thanks in advance
What limit are you talking about? Perhaps it would be a good idea if you stated exactly what the problem you are working on is! You have already been given the answer to the problem you stated.