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Chocolate bars and arithmetic progressions
These are two stories about Matthew, his father, chocolate bars and arithmetic progressions.
Story 1
One day, twelve years old Matthew was in a grocery store with his father.
Matthew saw the chocolate bars stacked on a shelf in a pile in a way
similar to that shown in the Figure 1.
- Dad, how many chocolate bars are there in the pile? - Matthew asked his father.
- How many bars are there at the base? - father asked in response.
- Where?
- In the bottom layer.
- Ten bars - answered Matthew.
- O, there are 55 chocolate bars in the pile - father calculated in three seconds.
- Father, how could you count it so fast? - Matthew asked.
- Dear Matthew, I learned it at school and recently refreshed
my memory reading the lesson Arithmetic progression in the site
www.Algebra.com .
You see, Matthew, - continued father, - there is one chocolate bar
in the uppermost layer. Then there are two bars in the next layer down,
then three bars in the next layer down, and so on till 10 bars in the
bottom layer. Hence, what you need to do is to sum the arithmetic
progression with the terms 1, 2, 3, ... , 10.
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Figure 1. The pile of chocolate bars (to the Story 1).
The stack is shown for 5 bars at the base only, not for 10.
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The lesson says that the sum of the first n natural numbers is equal to  .
So, I substituted n=10 into this formula and got the answer  =  =  .
- Thank you, dad, it is very instructive, - Matthew said.
Story 2
Another day, twelve years old Matthew was in the grocery store
with his father again. Matthew saw the chocolate bars stacked on a shelf
in a pile in a way similar to that shown in the Figure 2.
- Dad, how many chocolate bars are there in the pile? - Matthew asked his father.
- How many bars are there at the base? - father asked in response.
- Where?
- In the bottom layer.
- Ten bars - Matthew answered.
- O, there are 55 chocolate bars in the pile - father calculated in three seconds.
- Father, how could you count it so fast? - Matthew asked.
- Dear Matthew, do you remember we talked about another configuration
of the chocolate bars pile before? (See the Story 1). Here you have
the same arithmetic progression: one bar in the uppermost layer, then
two bars in the next layer down, then three bars in the next
layer, and so on till 10 bars in the bottom layer.
Since the sequence is the same, the sum is the same too.
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Figure 2. The pile of chocolate bars (to the Story 2).
The stack is shown for 5 bars at the base only, not for 10.
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But this time I can explain it differently, - father said.
Look in the Figure 3 on the right.
Imagine that I put 10 additional chocolate bars over the single
leftmost chocolate bar. They are shown in green in the Figure 3.
Note that these additional bars are all imaginary, and I do not
disturb existing bars in the right stack or in the entire pile.
In all I have 11 bars in this leftmost stack now, including one
real and ten imaginary, correct?
Next, I put imaginary chocolate bars over the two existing bars
to the second stack from the left. I put there 9 additional bars,
the exact amount of existing bars in the stack next to the last
one. They are shown in green in the second stack from the left
in the Figure 3. In all I have 11 bars in this stack now,
2 real and 9 imaginary, correct?
I will continue this procedure moving from the left to the right
and completing one stack after another. In the current i-th stack
I will put the additional bars in the amount equal to that of
existing bars in the i-th stack counted from the right.
To the last, 10-th stack I will put one additional imaginary bar,
according to the only one existing bar in the very left stack.
All the additional bars are shown in green color in the Figure 3.
The full number of bars (including existing real and added imaginary)
is obviously the constant quantity equal to 11 for all the stacks.
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Figure 3. The pile of chocolate bars (to the Story 2).
The stack is shown for 5 bars at the base only, not for 10.
The added imaginary bars are shown in green color.
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So, at the end I will have 11 bars, real and imaginary, in each of 10 stacks, totally 11*10 = 110 bars. Exactly half of them are real bars, the rest are imaginary.
Thus the number of the real bars is 11*10/2 = 55.
Do you understand, Matthew, that this is another way to calculate the sum of the first n natural numbers?
- Yes, I understand it, father. Thank you, - Matthew said.
- Thank you for asking. Hope you memorize this procedure.
My lessons on arithmetic progressions in this site are
- Arithmetic progressions
- The proofs of the formulas for arithmetic progressions
- Problems on arithmetic progressions
- Word problems on arithmetic progressions
- Chocolate bars and arithmetic progressions (this lesson)
- Free fall and arithmetic progressions
- Uniformly accelerated motions and arithmetic progressions
- Increments of a quadratic function form an arithmetic progression
- One characteristic property of arithmetic progressions
- Solved problems on arithmetic progressions
- Calculating partial sums of arithmetic progressions
- Finding number of terms of an arithmetic progression
- Inserting arithmetic means between given numbers
- Advanced problems on arithmetic progressions
- Interior angles of a polygon and Arithmetic progression
- Math Olympiad level problems on arithmetic progression
- Problems on arithmetic progressions solved MENTALLY
- Entertainment problems on arithmetic progressions
- Mathematical induction and arithmetic progressions
- Mathematical induction for sequences other than arithmetic or geometric
OVERVIEW of my lessons on arithmetic progressions with short annotations is in the lesson OVERVIEW of lessons on arithmetic progressions.
Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II.
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