Another difficult series question help
Show that there is exactly one positive integer n for which:
∑r^3+∑r=8∑r^2 (r=1 always) (n on top always)
I subbed in the appropriate formulas:
1/4n^2(n+1)^2 + 1/2n(n+1) = 8(1/6)n(n+1)(2n+1)
I don't know what to do next? Am i supposed to solve n if so how?
∑r^3+∑r=8∑r^2 (r=1 always) (n on top always)
I subbed in the appropriate formulas:
1/4n^2(n+1)^2 + 1/2n(n+1) = 8(1/6)n(n+1)(2n+1)
I don't know what to do next? Am i supposed to solve n if so how?
Original post by jarjarmonkey
Show that there is exactly one positive integer n for which:
∑r^3+∑r=8∑r^2 (r=1 always) (n on top always)
I subbed in the appropriate formulas:
1/4n^2(n+1)^2 + 1/2n(n+1) = 8(1/6)n(n+1)(2n+1)
I don't know what to do next? Am i supposed to solve n if so how?
∑r^3+∑r=8∑r^2 (r=1 always) (n on top always)
I subbed in the appropriate formulas:
1/4n^2(n+1)^2 + 1/2n(n+1) = 8(1/6)n(n+1)(2n+1)
I don't know what to do next? Am i supposed to solve n if so how?
Yes
Well, I would multiply by something to remove the fractions
Then divide through by the 2 common factors
Then solve the quadratic
Original post by TenOfThem
Yes
Well, I would multiply by something to remove the fractions
Then divide through by the 2 common factors
Then solve the quadratic
Well, I would multiply by something to remove the fractions
Then divide through by the 2 common factors
Then solve the quadratic
I multiplied by twelve to get 3n 6n and 16n what do u mean by divide through by two common factors?
Original post by jarjarmonkey
I multiplied by twelve to get 3n 6n and 16n what do u mean by divide through by two common factors?
There are 2 things that are common to each term
Original post by TenOfThem
There are 2 things that are common to each term
the only thing thats common is 1 though
Original post by jarjarmonkey
the only thing thats common is 1 though
Are you saying that you do not have
I did not check your working but I thought that was what you said
Original post by TenOfThem
Are you saying that you do not have
I did not check your working but I thought that was what you said
I did not check your working but I thought that was what you said
I have that but the only thing common between 3, 6, and 16 is 1.
Sorry if you mean something else and what your saying is really obvious.
Original post by jarjarmonkey
what your saying
You're.
There are algebraic terms to consider as well as numbers.
(edited 9 years ago)
Original post by Mr M
There are algebraic terms as well as numbers.
oh
Original post by jarjarmonkey
I have that but the only thing common between 3, 6, and 16 is 1.
Sorry if you mean something else and what your saying is really obvious.
Sorry if you mean something else and what your saying is really obvious.
I think it is obvious
For example, n is a common factor
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