**Sheri finds the Volume
of a Pyramid is 1/3 the Volume
of a Cube with the same base and height**

Sheri built 3 pyramids out of cm cubes and tried to fit them together to make a cube, but that didn't work.

1. Sheri found 3 pyramids in the Geoblock set that did make a cube. So the volume of the pyramid was 1/3 the volume of the cube with the same base and height.

2. Then Don asked Sheri to fill a plastic square
pyramid** **with water and
see how many it would take to fill the plastic cube. It took 3 pyramids with
water to fill the cube. So the volume of the pyramid was 1/3 the volume of the
cube with the same base and height. These plastic pieces are from the
"Volume Relationship Set" which was obtained from *Ideal
School Supply Company*.

3. Don asked Sheri to find the ratio of the volume of a pyramid made with cm cubes to the volume of a cube also made of cm cubes, with the same base and height. The volume of the pyramids were the sum of squares, and the cubes were each a cube number. The picture of the wood pieces is shown below

Sheri's work to find the sum of the squares, the cubes, and the ratios of the volumes as a fraction and a decimal is shown below. Sheri worked very hard on these!

**Fine work Sheri!**

Using *Mathematica*, Don and Sheri did the first 4 below;
each gives the input, then the second line is the output. Then Don did the rest
after Sheri left, but he showed her what he did later.

**1.** below says: Find the sum of n^{2}, from
n=1 to n=3; in other words, add 1^{2} + 2^{2} + 3^{2} =
1 + 4 + 9 = 14

^{1.}^{
}**
**

14

**2.**
below says: Find the sum of n^{2}, from n=1 to n=100; in other words,
add 1^{2} + 2^{2} + 3^{2} + .. +100^{2
}= 1 + 4 + 9 +.. + 10000 =
338350

^{2}^{.}^{
}**
**

338350

3.**
**100^3

1000000

^{4}^{.}^{
}

0.33835

**2. **above shows that going
to 100 layers of the pyramid the volume is 338350, **3**.
above shows
the volume of the
cube of side 100 is 1000000. **4**.
above shows the ratio of the Volume of the pyramid/ Volume of the cube Sheri
found above 0.33835. The command in **4.** tells the computer to divide
338350 by 1000000 and give the answer to 10 digits (it doesn't give all 10
digits if there are zeros on the right). Don made sure Sheri could get the
decimal for 1/3 = 0.33333... (and he checked her division for the other ratios).

The ratios Sheri was getting form an **infinite
sequence**

1,
5/8, 14/27, 30/64, 55/125,
91/216, .. , 338350/1000000,... or as decimals

1, 0.625, 0.518..., 0.468,
0.440, 0.421,..,
0.33835,...

This sequence is a **slowly converging**
sequence; it took 100 terms to reach 0.33835 and
is getting closer to 0.33333... = 1/3. Don used *Mathematica* to show
that the sequence really does go to 1/3.

In **5.**
below Don used *Mathematica *to find the sum of the first x squares.

^{5}^{.}^{
}**
**

**
**

In

^{6.}^{
}**
**

**
**

In
**7. **below Don** **used
*Mathematica *to find the limit* *of the function above as x goes to
Infinity, which is **1/3**.**
**

^{7.}^{
}**
**

**
**

Don shows below how he found the rule for the sum of the first x square numbers given in

a

In a**x**^{3} + b**x**^{2} + c**x**
+ d = **y **, if we put** ****x**=1 and **y**=1, then **x**=
2 and **y**=5, then **x**= 3 and **y**=14, then **x**= 4 and **y**=30,
we get a system of 4 equations in 4 variables a, b, c, and d. **The
equation number is in red**.

Putting in

x=1, **Eq1 = **a
+ b + c +
d = 1

x=2, **Eq****2
=** 8a + 4b + 2c
+ d = 5

x=3, **Eq****3
=** 27a + 9b + 3c + d =
14

x=4, **Eq****4
= **64a + 16b + 4c + d =
30

To eliminate d:

**Eq2-****Eq****1=****Eq5=
**7a + 3b + c = 4

**Eq3-****Eq****2=****Eq****6=**19a
+ 5b + c = 9

**Eq****4-****Eq****3=****Eq****7=**37a
+ 7b + c =16

To eliminate c:

**Eq****6-****Eq****5=****Eq8=
**12a + 2b = 5

Eq**7-****Eq****6=****Eq9=
**18a + 2b = 7

To eliminate b

**Eq****9-****Eq****8=****Eq10=
**6a=2, so **a = 1/3**

To solve for b, put **a = 1/3 **in
**Eq8**:

12* 1/3 +2b =5

4 + 2b = 5

**b = 1/2**.

To solve for c, put **a = 1/3 **and
**b = 1/2 **into **Eq****5
**:

7*1/3 + 3*1/2 + c = 4 Multiply both sides by 6

14 + 9 + 6c = 24

6c = 1

**c = 1/6**

To solve for d, put **a = 1/3
**and **b = 1/2 **and
**c = 1/6 **into **Eq1**,
and we get

1/3 + 1/2 + 1/6 + d = 1, and from this

1 + d = 1, so

**d = 0**

If we now put **a = 1/3**, **b
= 1/2**, **c = 1/6**, and **d
= 0** in the general case

a**x**^{3} + b**x**^{2} + c**x** + d = **y **, we
get:

1/3**x**^{3 }+ 1/2**x**^{2} + 1/6**x**
= **y **factoring 1/6**x **from each term, we get

1/6**x** (2**x**^{2}
+ 3**x **+ 1) and factoring the trinomial, we get

1/6**x** (2**x** + 1)(**x
**+ 1) which is the same as **5.** above.

Now if we divide 1/3**x**^{3 }+
1/2**x**^{2} + 1/6**x **by** x**^{3
}, we get 1/3 + 1/(2x) + 1/(6**x**^{2}) which is the same
as **6.** above.

If we take the limit of 1/3 + 1/(2x) + 1/(6**x**^{2}) as x
->Infinity, the two terms on the right go to zero, so we get **1/3**,
which agrees with **7.** above, and the wooden pieces, and the water filled
pyramid and cube, and Sheri's work with the wooden pieces!

Mathman home