Assume that :
x = (n^2-n)/2
y = ((n+1)^2)-(n+1))/2
n is an integer and n ≥ 0
Examples =
3(0) + 1 = 1
3(1) + 3 = 6
3(3) + 6 = 15
and so on.
4 Answers
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1
Welcome to StackExchange!
This can be proven by doing this:
Multiplying x's formula by 3 and,
Rewriting the formula for y since $(n+1)^2-(n+1)=n(n+1)=n^2+n$,
Writing them in one fraction since their denominators are the same,
We get $3n^2-3n+n^2+n\over2$$=$$2n^2-2n\over2$=$n^2-n\over2$$=$$n(n+1)\over2$ which is the main formula for triangular numbers.
Also, I believe this question is more fitting for math.stackexchange.com
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1$\begingroup$ Although the content is fine, this doesn't seem to involve any Mathematica code $\endgroup$MarcoB– MarcoB2023-05-12 16:31:35 +00:00Commented May 12, 2023 at 16:31
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Triangle numbers have form B = 1/2 m (1 + m). Now
x = (n^2 - n)/2;
y = ((n + 1)^2 - (n + 1))/2;
A = Assuming[Element[n, Integers] && n <= 0, Simplify[3*x + y]]
So we just need to show that A is subset of B.
Reduce[(A == B) && n >= 0 && m >= 0, {n, m}, Integers]
So the above says that only for integer values of n==(m + 1)/2 then n numbers are triangles.
rel = (m + 1)/2
Cases[Table[rel, {m, 0, 10}], x_ /; IntegerQ[x]]
A /. n -> %
The above are triangle numbers. They are subset of the numbers generated by B
Table[B, {m, 0, 10}]
did you mean to ask this in the mathematics group? This is Mathematica forum with an a at the end instead of s.
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May as well add another approach...
Block[
{x = (n^2 - n)/2,
y = ((n + 1)^2 - (n + 1))/2},
Solve[3 x + y == Binomial[k, 2], k]
]
(* {{k -> 1 - 2 n}, {k -> 2 n}} *)
The question is clearly not about Mathematica, but since it's attracted so many Mma answers already, it's probably too late to migrate it.
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Here is one easy approach
T[n_] := (n^2-n)/2;
3*T[n] + T[n+1] == T[2*n] //Simplify
(* True *)
y. When you remove it evaluate:3 x + y // Simplifyand see if the sequence function generates triangular numbers. Thanks. $\endgroup$