# Rewriting a sum

1. Nov 8, 2008

### jorgen

Hi all,

I have the following sum

$$\sum n - n_1 + 1$$

which I split up in two independent sums

$$\sum_{n_1=0}^N n + 1 - \sum_{n_1=0}^N n_1$$

the last sum can be written as

$$0.5*n(n+1)$$

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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 8, 2008

### CompuChip

Unless you are holding something back,
$$\sum_{n_1=0}^N n + 1 = (N + 1)(n + 1)$$

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

3. Nov 8, 2008

### jorgen

thanks for the reply - my problem is understanding the first summation

$$(N+1)(n+1)$$

how is small n to be interpreted?

Thanks in advance any hints appreciated.

4. Nov 8, 2008

### HallsofIvy

Staff Emeritus
However YOU mean it! You wrote
$$\sum_{n1= 0}^N (n+1)$$
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 $\sum n- n1+ 1$ would be interpreted as
[tex]\sum_{n=0}^N n- n1+ 1[/itex]
where n1 is some fixed number.

5. Nov 8, 2008

### jorgen

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?