Looking for patterns

**darksoulzero** · July 17th 2011, 12:57 PM

Hey guys, I have a question where the book asks me to produce a pattern after observing the difference in the changing numbers. This completely stumped me. Here's the question:

The numbers 1, 8, 27, and 64 are the first 4 cubes.

a) Find the sum of the first 2 cubes. ans. 9
b) Find the sum of the first 3 cubes. ans. 36
c) Find the sum of the first 4 cubes. ans. 100
d) Describe the pattern in the sums.
e) Use the pattern to find the sum of the first 9 cubes.

So I determined the sums but I couldn't develop an expression that explains the pattern. I already looked at the resulting answer, but I'm not even sure what the hell I'm supposed to do to even get close to the answer. Am I supposed to use trial and error?

**skeeter** · July 17th 2011, 01:05 PM

Originally Posted by darksoulzero

Hey guys, I have a question where the book asks me to produce a pattern after observing the difference in the changing numbers. This completely stumped me. Here's the question:

The numbers 1, 8, 27, and 64 are the first 4 cubes.

a) Find the sum of the first 2 cubes. 9
b) Find the sum of the first 3 cubes. 36
c) Find the sum of the first 4 cudes. 100
d) Describe the pattern in the sums.
e) Use the pattern to find the sum of the first 9 cubes.

So I determined the sums but I couldn't develop an expression that explains the pattern. I already looked at the resulting answer, but I'm not even sure what the hell I'm supposed to do to even get close to the answer. Am I supposed to use trial and error?

$3^2 , 6^2 , 10^2 , ...$

would you agree the next term is $15^2$ ?

**darksoulzero** · July 17th 2011, 04:27 PM

Originally Posted by skeeter

$3^2 , 6^2 , 10^2 , ...$

would you agree the next term is $15^2$ ?

If the pattern is what I think it is, then yeah, it's $15^2$ . Is the next term $21^2$ ?

**skeeter** · July 17th 2011, 04:36 PM

Originally Posted by darksoulzero

If the pattern is what I think it is, then yeah, it's $15^2$ . Is the next term $21^2$ ?

now that you've determined a pattern, what is the sum ...

$1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = \sum_{k=1}^n k^3$

?

**Plato** · July 17th 2011, 04:38 PM

Originally Posted by darksoulzero

If the pattern is what I think it is, then yeah, it's $15^2$ . Is the next term $21^2$ ?

Can you use induction to prove this?:

$\sum\limits_{k = 1}^N {k^3 } = \left( {\frac{{N^2 + N}}{2}} \right)^2,~~N\ge 1$

**darksoulzero** · July 17th 2011, 05:49 PM

Originally Posted by skeeter

now that you've determined a pattern, what is the sum ...

$1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = \sum_{k=1}^n k^3$

?

Ohh, I see now. Hmm, but the pattern at the back of my book is described as
$\frac{n^2(n + 1)^2}{4}$

And I don't understand what that means.

Originally Posted by Plato

Can you use induction to prove this?:

$\sum\limits_{k = 1}^N {k^3 } = \left( {\frac{{N^2 + N}}{2}} \right)^2,~~N\ge 1$

Hmm, I don't think so. I tried to,but then I realized that I don't know how to prove something. Can I prove it's true by showing that the numbers 1-4 correlate to the numbers of the first 4 cubes? Or does it correlate to the sum of the numbers before N including the Nth term?

**skeeter** · July 17th 2011, 06:18 PM

here is the sequence of partial sums (w/o the squares, ignore that for the meantime)

$1, 3, 6, 10, 15, 21, ...$

double this sequence ...

$2, 6, 12, 20, 30, 42, ...$

note how each term of the sequence factors (see the pattern?)

$(1 \cdot 2) , (2 \cdot 3) , (3 \cdot 4), (4 \cdot 5) , (5 \cdot 6) , (6 \cdot 7) , ...$

"un" double ...

$\frac{1}{2} \left[(1 \cdot 2) , (2 \cdot 3) , (3 \cdot 4) , (4 \cdot 5) , (5 \cdot 6) , (6 \cdot 7) , ...\right]$

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