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August 11th, 2017, 09:51 AM   #1
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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.

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Last edited by skipjack; August 11th, 2017 at 12:03 PM.
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August 11th, 2017, 10:34 AM   #2
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They are 'triangular numbers'. They are of the form $\frac{n(n+1)}{2}$.
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August 11th, 2017, 10:44 AM   #3
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Thank you !
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August 11th, 2017, 12:14 PM   #4
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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).
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August 12th, 2017, 12:02 AM   #5
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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 12:08 AM.
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