# Need help understanding summation notation problem...

• Nov 23rd 2006, 01:50 AM
ergoSum
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.
• Nov 23rd 2006, 02:13 AM
topsquark
Quote:

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
• Nov 23rd 2006, 03:51 AM
CaptainBlack
Quote:

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.

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

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

RonL