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

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Homework Help Overview

The discussion revolves around deriving the formula 0.5(n+1)(n+2) from a summation problem involving the expression \sum n - n_1 + 1. Participants are exploring the interpretation and manipulation of summations in the context of this formula.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to split the original summation into two parts and are discussing how to rewrite the first summation. There are questions about the interpretation of variables, particularly the small 'n' and the index 'n_1'.

Discussion Status

Some participants have provided interpretations and potential simplifications of the summation, while others are seeking clarification on the notation and the meaning of the variables involved. There is an ongoing exploration of how to arrive at the desired formula without a clear consensus on the approach.

Contextual Notes

There are indications of confusion regarding the notation used, particularly with the variable 'n_1' and its role in the summation. Additionally, the original poster's intent regarding limits and the relationship between 'n' and 'N' is not fully articulated, leading to further questions.

jorgen
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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
 
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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.
 

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