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consecutive integers raised to power of 6, such as X(n)^6=X(
- Thread starter mgerace
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Re: consecutive integers raised to a power
mgerace said:Is there an algebraic expression that describes the relationship between consecutive integers to the power of 6, such as X(n)^6 = X(n-1)^6 +???????
What does X indicate here?
Re: consecutive integers raised to a power
X is an integer. I'm trying to figure out if X(n)^6 can be explained by the previous integer and additional terms, that is, X(n-1)^6 + terms.
For example, the relationship between consecutive integers to the power of 5 is easy to figure out.
It's X(n)^5 = X(n-1)^5 + 5[X(n-1)^3]*[X(n)] + 5[X(n-1)]*[X(n)^2] + 1
No, I don't know how to expand (a+b)^6, but this seems like a good suggestion.
Thanks.
X is an integer. I'm trying to figure out if X(n)^6 can be explained by the previous integer and additional terms, that is, X(n-1)^6 + terms.
For example, the relationship between consecutive integers to the power of 5 is easy to figure out.
It's X(n)^5 = X(n-1)^5 + 5[X(n-1)^3]*[X(n)] + 5[X(n-1)]*[X(n)^2] + 1
No, I don't know how to expand (a+b)^6, but this seems like a good suggestion.
Thanks.
Re: consecutive integers raised to a power
Just to let you know, I just expanded (a+b)^6 and does not seem to fit the problem. I also compared the (a+b) expansion to the powers of 2, 3, 4 and 5 for the actual algebraic expressions for the consecutive integers at those powers respectively. They do not resemble te solutions.
For example, (a+b)^4 does not seem to describe X(n)^4 = X(n-1)^4 + 4[X(n-1)^2]*[X(n)] + 2[X(n)^2] -1 which is the solution for consecutive integers raised to the power of 4.
Just to let you know, I just expanded (a+b)^6 and does not seem to fit the problem. I also compared the (a+b) expansion to the powers of 2, 3, 4 and 5 for the actual algebraic expressions for the consecutive integers at those powers respectively. They do not resemble te solutions.
For example, (a+b)^4 does not seem to describe X(n)^4 = X(n-1)^4 + 4[X(n-1)^2]*[X(n)] + 2[X(n)^2] -1 which is the solution for consecutive integers raised to the power of 4.
Re: consecutive integers raised to a power
I admit to being very confused.
Consecutive integers to the power of 6???
Consecutive integers might be represented as follows:
let x = the first integer
Then x + 1 = the next consecutive integer
And x + 2 = the consecutive integer after THAT one
Now...what in the world do you mean by "consecutive integers to the power of 6?"
If you have a series of consecutive integers, like 5, 6, 7, 8 or x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), then just what does
"consecutive integers to the power of 6" mean??
If you can provide some more specific information about the problem you're trying to solve, and show us what work you've done, we MAY be able to assist.
mgerace said:Is there an algebraic expression that describes the relationship between consecutive integers to the power of 6, such as X(n)^6 = X(n-1)^6 +???????
Thanks.
I admit to being very confused.
Consecutive integers to the power of 6???
Consecutive integers might be represented as follows:
let x = the first integer
Then x + 1 = the next consecutive integer
And x + 2 = the consecutive integer after THAT one
Now...what in the world do you mean by "consecutive integers to the power of 6?"
If you have a series of consecutive integers, like 5, 6, 7, 8 or x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5), then just what does
"consecutive integers to the power of 6" mean??
If you can provide some more specific information about the problem you're trying to solve, and show us what work you've done, we MAY be able to assist.