eScribe Poste Mauricio Gomelo and Abraham Faraon

Scribe Post
Mauricio Gomez and Abraham Farah
In today’s lesson we learned a more advanced concept of Gauss’s technique. All problems done in class were about arithmetic series, but, not all series are arithmetic. Also remember that Sn means “the sum of all the first n terms of the series”

Our daily POD was about arithmetic series.
POD 5/7 : A pile of logs has 30 logs in the bottom row. Each higher row has one fewer ending with 9 logs in the top row. How many logs are in the pile?

Even though there can be many ways to solve this problem, we discussed two of them in class. They involved the following: The bottom pile has 30 logs and each continuous pile decreases by one log, till reaching nine logs. Therefore, the sequence goes from 9 to 30. However, in order to make this problem work properly we also have to take into consideration that there are 8 series integers missing from 1 to 8, since the sequence starts at 9. We can apply Gauss Technique using this consideration.
S(30)- S(8)= S(5)
Sn= n/2(n+1)
S(30)=30/2(30+1)=465
S(8)=8/2(8+1)=36
465-36= 429

OR

From 9 to 30 are 22 integers including 9 and 30
Those integers make 11 pairs, or 22/2 = 11
The sum of the pair of each opposite integer is 39.
39 * 11 = 429

Both processes work.

During the rest of the class we did two more series problems.
1. Find S(5) for the following arithmetic series 5+10+15+20+25 The next two terms are obviously 20 and 25 The sum of opposite terms that are paired is 30. For instance 5+25=30, 10+20=30, etc. Since there are no even pairs you do the following:
30 * 5/2 = 75
2. Find S(5) for the following arithmetic series: 35+40+45+50+55
The sum of the pairs is 90
90 * 5/2=225
Gauss’s method works for the 1st n terms of any arithmetic series
Gauss’s updated formula
Sn= n/2 (T(1) + T(n))
This week’s hw so far was posted last Thursday: pg. 489 #1-8
Next Scribe Post= Cristina Franco, Mari Guillen