algebra -number pattern
- Thread starter elmer
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
I will never, EVER, let this go. It is a technical error. It is NOT a reasonable question to suggest to the student that there is one and only one proper response to this sort of question. There are INFINITELY MANY correct responses. Some have attempted to define a "simplest" response, but these attempts are feeble and unsatisfactory.
One should find "A" next value and then be required to explain how it was obtained. These is no excuse for insisting on "THE" correct value. If you can argue it and explain it, it should be considered a correct answer.
Having said that, there are a few simple tools to find some pattern. One very popular is called a "difference". Just subtract successive valeus and see what you get.
8-1 = 7
27-8 = 19
64-27 = 37
That doesn't look like much, does it. Do it again.
19-7 = 12
37-19 = 18
This says something to me. You may not see it. I'll leave it at that for this "difference" exploration. We rather ran out of values. Trust me on this. This difference method will help you solve these.
Anyway, that was only a demonstration. In this case, there is a quick solution:
\(\displaystyle 1^{3} = 1\)
\(\displaystyle 2^{3} = 8\)
\(\displaystyle 3^{3} = 27\)
\(\displaystyle 4^{3} = 64\)
One should find "A" next value and then be required to explain how it was obtained. These is no excuse for insisting on "THE" correct value. If you can argue it and explain it, it should be considered a correct answer.
Having said that, there are a few simple tools to find some pattern. One very popular is called a "difference". Just subtract successive valeus and see what you get.
8-1 = 7
27-8 = 19
64-27 = 37
That doesn't look like much, does it. Do it again.
19-7 = 12
37-19 = 18
This says something to me. You may not see it. I'll leave it at that for this "difference" exploration. We rather ran out of values. Trust me on this. This difference method will help you solve these.
Anyway, that was only a demonstration. In this case, there is a quick solution:
\(\displaystyle 1^{3} = 1\)
\(\displaystyle 2^{3} = 8\)
\(\displaystyle 3^{3} = 27\)
\(\displaystyle 4^{3} = 64\)
its a very simple number series. it says 1 , 8 , 27 , 64 and so on.
1= 1 to the power of 3 as in 1x1x1=1
8= 2 to the power of 3 as in 2x2x2=8
27= 3 to the power of 3 as in 3x3x3=27
64= 4 to the power of 3 as in 4x4x4=64
the next number in the series will be
125= 5 to the power of 3 as in 5x5x5=125
hope my explanation was clear to you. math especially algebra is a very interesting subject if you get the basic concepts right , am sure you will start enjoying math.
1= 1 to the power of 3 as in 1x1x1=1
8= 2 to the power of 3 as in 2x2x2=8
27= 3 to the power of 3 as in 3x3x3=27
64= 4 to the power of 3 as in 4x4x4=64
the next number in the series will be
125= 5 to the power of 3 as in 5x5x5=125
hope my explanation was clear to you. math especially algebra is a very interesting subject if you get the basic concepts right , am sure you will start enjoying math.
krishnan said:
krishnan,
no, there are an infinite number of possibilities.
Here is just one of the rules:
N(x) = (91/24)x^4 - (443/12)x^3 + (3185/24)x^2 - (2275/12)x + 91
Here, N(5) = 216, not 125.
---------------------------------------------------------------------------------------------
Here's another example:
1, 4, ...
What is the next number in the sequence?
Arithmetic ---> 7
Alternating ---> 1
Perfect squares ---> 9
Geometric----> 16
n^n---> 1^1, 2^2, 3^3 ---> 27
tkhunny wrote:
Having said that, there are a few simple tools to find some pattern. One very popular is called a "difference". Just subtract successive valeus and see what you get.
8-1 = 7
27-8 = 19
64-27 = 37
That doesn't look like much, does it. Do it again.
19-7 = 12
37-19 = 18
This says something to me. You may not see it. I'll leave it at that for this "difference" exploration. We rather ran out of values. Trust me on this. This difference method will help you solve these.
Yes, the method of finite differences would clearly nail down a unique solution.
Given:
n.....1.....2.....3.....4...
N.....1.....8....27...64...
Diff.....7....19...37...
Diff.......12....18
Diff...........6
Unfortunately, the length of the given sequence does not allow us to conclude that we have a finite difference sequence. But what if it were?
Lets assume that the third difference was a constant 6.
n.....1.....2.....3.....4.....5.....6...
N.....1.....8....27...64...125..216...
Diff.....7....19....37...61...91...
Diff.......12....18...24...30...
Diff...........6.....6.....6...
With the third differences constant at 6, this defines the expression for the nth term of the form N = an^3 + bn^2 + cn + d.
Using the given data, we can write
a + b + c + d = 1
8a + 4b + 2c + d = 8
27a + 9b + 3c + d = 27
64a + 16b + 4c + d = 64
Solving, a = 1, b = 0, c = 0 and d = 0 leaving us with N = n^3
Having said that, there are a few simple tools to find some pattern. One very popular is called a "difference". Just subtract successive valeus and see what you get.
8-1 = 7
27-8 = 19
64-27 = 37
That doesn't look like much, does it. Do it again.
19-7 = 12
37-19 = 18
This says something to me. You may not see it. I'll leave it at that for this "difference" exploration. We rather ran out of values. Trust me on this. This difference method will help you solve these.
Yes, the method of finite differences would clearly nail down a unique solution.
Given:
n.....1.....2.....3.....4...
N.....1.....8....27...64...
Diff.....7....19...37...
Diff.......12....18
Diff...........6
Unfortunately, the length of the given sequence does not allow us to conclude that we have a finite difference sequence. But what if it were?
Lets assume that the third difference was a constant 6.
n.....1.....2.....3.....4.....5.....6...
N.....1.....8....27...64...125..216...
Diff.....7....19....37...61...91...
Diff.......12....18...24...30...
Diff...........6.....6.....6...
With the third differences constant at 6, this defines the expression for the nth term of the form N = an^3 + bn^2 + cn + d.
Using the given data, we can write
a + b + c + d = 1
8a + 4b + 2c + d = 8
27a + 9b + 3c + d = 27
64a + 16b + 4c + d = 64
Solving, a = 1, b = 0, c = 0 and d = 0 leaving us with N = n^3
mmm4444bot
Super Moderator
- Joined
- Oct 6, 2005
- Messages
- 10,902
In this thread, we might limit our considerations to topics in arithmetic.