develop a formula for the solution of a recurrence relation of the form
a[n] = a[n-1] + m, a[1] = m

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Clarification of Question by percy227-ga on 15 Dec 2003 17:51 PST

need all the work

Hello percy!
This problem can be solved in the following way:

We know that

a[n] = a[n-1] + m   (1)
a[1] = m            (2)

From equation (1) we also have that

a[n-1] = a[n-2] + m, as long as n>=3

So plugging this again into equation (1) gives:

a[n] = a[n-2] + m + m

We can replace again a[n-2] using equation (1). Notice that we can
continue replacing this term up to n-2 times. For example, take a[4].
We have

a[4] = a[3] + m
a[4] = a[2] + m + m
a[4] = a[1] + m + m + m

And now we can't replace a[1] by something that depends on a[0],
because a[0] does not exist. So we were able to replace the term in
the right hand size 4-2=2 times.Notice also that each time we replace
this term, we have to sum another 'm'. Therefore, we have that:

a[n] = a[1] + m*(n-2) + m

This is so because we replace the term in the right hand side (a[n-1],
then a[n-2], etc) n-2 times, leaving us with the a[1], the original +m
and (n-2) times m. Finally, using the fact that a[1] = m, we have:

a[n] = m + m*(n-2) + m
a[n] = m*n

For more information on discrete structures, you may want to check the link below.

Introduction to Discrete Structures
http://www.cs.odu.edu/~toida/nerzic/content/web_course.html


Google search strategy
discrete structures
://www.google.com.ar/search?q=discrete+structures&ie=UTF-8&oe=UTF-8&hl=es&meta=


I hope this helps! If you have any doubt regarding my answer, please
request a clarification before rating it. Otherwise I await your
rating and final comments.

Best wishes!
elmarto

i understand fully what is done here
thanks