My Math Forum trying to find a name for this simple sequence?
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 August 11th, 2017, 10:51 AM #1 Newbie   Joined: Aug 2017 From: san jose, ca Posts: 5 Thanks: 0 trying to find a name for this simple sequence? n=1: 1 n=2: 1 + 2 n=3: 1 + 2 + 3 n=k: 1 + 2 + 3 + ... + k It has been long since I am in college and I looked at factorial and binomial only to find I grossly missed. Thanks. Last edited by skipjack; August 11th, 2017 at 01:03 PM.
 August 11th, 2017, 11:34 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond They are 'triangular numbers'. They are of the form $\frac{n(n+1)}{2}$. Thanks from Country Boy
 August 11th, 2017, 11:44 AM #3 Newbie   Joined: Aug 2017 From: san jose, ca Posts: 5 Thanks: 0 Thank you !
 August 11th, 2017, 01:14 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,865 Thanks: 1833 Had you asked us to name this sequence: n=1: 1 n=2: 1 + 3 n=3: 1 + 3 + 5 n=k: 1 + 3 + 5 + ... + (2k-1) we would have answered that these numbers are the positive perfect squares. Each sequence is a particular example of a quadratic sequence (because the value of the kth member of the sequence is given by a quadratic function of k).
 August 12th, 2017, 01:02 AM #5 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 And, more in general the sequence for all Powers: 1^n=1 2^n=1+(2^n-(2-1)^n) ... x^n=1+....+(x^n-(x-1)^n) or: $$A^n=\sum_{x=1}^{A}(x^n-(x-1)^n)$$ Known as telescoping sum properties. I use $x$ instead of "i" since all this becomes very interesting once sticked on a Cartesian Plane: Here the most simple case for $n=2$: And playing with the binomial develope / sum and fractions properties you can have back a Power of a Rational too. Here an example for n=3: Calling: $M_n= {[ x^n-(x-1)^n]}$ it becomes for $n=3$: $M_3= {( 3x^2-3x+1 )}$ Than: $A^3 =\sum_{x=1}^{A}{( 3x^2-3x+1)}$ We can divide all the terms of our Sum by $K^3$ , remembering that if we want to left unchanged the result, we have to multiply the Upper limit by $K$ so we have: $A^3 = \frac {K^3*A^3} {K^3} = \sum_{x=1}^{A*K}{( 3x^2/{K^3}-3x/{K^3}+1/{K^3})}$ Now we can call: $x= x/K$ , so changing $x$ with $x= x*K$, if we respect the following conditions: a) if and only if $K$ is a Factor of $A$, or perfectly divide $A$ - the Upper limit becomes: $K*A/K = A$ (with $K,A\in N*$ ) - the Lower limit becomes: $x=x/K$ so $x= 1$ becomes $x=1/K$ so: $A^3 = \sum_{x=1/K}^{A}{( 3(x*K)^2/{K^3}-3(x*K)/{K^3}+1/{K^3}) }$ Now we can simplify to have our new Step Sum, that moves Step $1/K$ from $1/K$ to $A$, so the Index $x$ will be $x =1/K, 2/K,3/K.... A$: $A^3 = \sum_{x=1/K}^{A}{\left( \frac {3x^2}{K}- \frac {3x}{K^2}+\frac{1}{K^3}\right)}$ for any $A\in \mathbb{Q}$ And pulling the Rational for $K\to\infty$ you discover the Integral: $$A^n=\int_{x=0}^{A} nx^{n-1} dx$$ for any $A\in \mathbb{R}$ Before share this with your Prof. remember is not yet accepted by any math "authority", still if it was mine 10 years old trick Last edited by complicatemodulus; August 12th, 2017 at 01:08 AM.

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