# Thread: Need help understanding summation notation problem...

1. ## Need help understanding summation notation problem...

So i'm doing some problems in my discrete math book. I ran into this little problem that has a solution in the appendix which I simply do not understand.

"Write each of 32-41 using summation or product notation.

32.
.
.
40. n + (n - 1) + (n - 2) + ... + 1
41."

The answer in the appendix says:

(n - i) where i = 0 is the lower limit and n-1 is the upper limit

I don't get this; that '... + 1' in the problem is throwing me off.

If you understand this, would you mind explaining this summation form to me.

Thank you.

2. Originally Posted by ergoSum
So i'm doing some problems in my discrete math book. I ran into this little problem that has a solution in the appendix which I simply do not understand.

"Write each of 32-41 using summation or product notation.

32.
.
.
40. n + (n - 1) + (n - 2) + ... + 1
41."

The answer in the appendix says:

(n - i) where i = 0 is the lower limit and n-1 is the upper limit

I don't get this; that '... + 1' in the problem is throwing me off.

If you understand this, would you mind explaining this summation form to me.

Thank you.
What is the value for n - i when i = 0? It's n.
What is the value for n - i when i = n - 1? It's n - (n - 1) = n - n + 1 = 1.

So the summation runs over all values from n down to 1.

-Dan

3. Originally Posted by ergoSum
So i'm doing some problems in my discrete math book. I ran into this little problem that has a solution in the appendix which I simply do not understand.

"Write each of 32-41 using summation or product notation.

32.
.
.
40. n + (n - 1) + (n - 2) + ... + 1
41."

The answer in the appendix says:

(n - i) where i = 0 is the lower limit and n-1 is the upper limit

I don't get this; that '... + 1' in the problem is throwing me off.

If you understand this, would you mind explaining this summation form to me.

Thank you.
$
n + (n - 1) + (n - 2) + ... + 1=\sum_{i=0}^{n-1} (n-i)
$

What the right hand side means is you add all the terms $(n-i)$ for $i=0,..,\ i=(n-1)$.
These are: when $i=0,\ n$, when $i=1,\ n-1$, ..., when $i=n-1,\ 1$
If you write this out in full you will find it equals the left hand side.

RonL