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Number Theory Express numbers, raised to an exponent, for sum of consecutive numbers.

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

Not long ago, I did a study on numbers, raised to an exponent. I noticed that a "pattern" remained, and I could find a general formula.

Let: [tex]a , n, k \in{N}[/tex], when "a" is odd number:

I define: [tex]a _1, \ldots , a_k \in{N}[/tex], as the consecutive addends, such that:

Let: [tex]a , n, k \in{N}[/tex], when "a" is even number:

I define: [tex]a _1, \ldots , a_k \in{Q}[/tex], as the addends, such that: [tex]a_{i+1}-a_i=1[/tex].


Two cases are presented to us:

1º) If “n”= even:

[tex]a_1= \dfrac{a^{\frac{n}{2}}+1}{2}[/tex]

[tex]a_k= \dfrac{3a^{\frac{n}{2}}-1}{2}[/tex]

[tex]k=a^{\frac{n}{2}}[/tex]

Demonstration:

[tex]a^n=\dfrac{\dfrac{a^{\frac{n}{2}}+1}{2}+\dfrac{3a^{\frac{n}{2}}-1}{2}}{2}a^{\frac{n}{2}}=[/tex]

[tex]=\dfrac{\dfrac{4a^{\frac{n}{2}}}{2}}{2}a^{\frac{n}{2}}= a^n[/tex]


2º) If “n” =odd:

[tex]a_1= \dfrac{a^{\frac{n+1}{2}}-2a^{\frac{n-1}{2}}+1}{2}[/tex]

[tex]a_k= \dfrac{a^{\frac{n+1}{2}}+2a^{\frac{n-1}{2}}-1}{2}[/tex]

[tex]k=2a^{\frac{n-1}{2}}[/tex]

Demonstration:

[tex]a^n=\dfrac{(\dfrac{a^{\frac{n+1}{2}}-2a^{\frac{n-1}{2}}+1}{2})+{(\dfrac{a^{\frac{n+1}{2}}+2a^{\frac{n-1}{2}}-1}{2})}}{2}2a^{\frac{n-1}{2}}=[/tex]

[tex]=\dfrac{2a^{\frac{n+1}{2}}}{4}2a^{\frac{n-1}{2}}=a^n[/tex]

Examples:

1º) n=6, a=5

[tex]a_1=\dfrac{5^{\frac{6}{2}}+1}{2}=63[/tex]

[tex]a_k=\dfrac{3*5^{\frac{6}{2}}-1}{2}=187[/tex]

[tex]k=5^{\frac{6}{2}}=125[/tex]

[tex]5^6=63+64+65+...+186+187=\dfrac{63+187}{2}125=15625[/tex]

2º) n=7. a=4

[tex]a_1=\dfrac{4^{\frac{7+1}{2}}-2*4^{\frac{7-1}{2}}+1}{2}=64.5[/tex]

[tex]a_k=\dfrac{4^{\frac{7+1}{2}}+2*4^{\frac{7-1}{2}}-1}{2}=191.5[/tex]

[tex]k=2*4^{\frac{7-1}{2}}=128[/tex]

[tex]4^7=64.5+65.5+66.5+...+190.5+191.5=\dfrac{64.5+191.5}{2}128=16384[/tex]


Note: I do not know if it will be "invented". If it is not, I could call it: "mente oscura" Theorem.(Rofl)

Regards.