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

mente oscura

Well-known member
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: $$a , n, k \in{N}$$, when "a" is odd number:

I define: $$a _1, \ldots , a_k \in{N}$$, as the consecutive addends, such that:

Let: $$a , n, k \in{N}$$, when "a" is even number:

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

Two cases are presented to us:

1º) If “n”= even:

$$a_1= \dfrac{a^{\frac{n}{2}}+1}{2}$$

$$a_k= \dfrac{3a^{\frac{n}{2}}-1}{2}$$

$$k=a^{\frac{n}{2}}$$

Demonstration:

$$a^n=\dfrac{\dfrac{a^{\frac{n}{2}}+1}{2}+\dfrac{3a^{\frac{n}{2}}-1}{2}}{2}a^{\frac{n}{2}}=$$

$$=\dfrac{\dfrac{4a^{\frac{n}{2}}}{2}}{2}a^{\frac{n}{2}}= a^n$$

2º) If “n” =odd:

$$a_1= \dfrac{a^{\frac{n+1}{2}}-2a^{\frac{n-1}{2}}+1}{2}$$

$$a_k= \dfrac{a^{\frac{n+1}{2}}+2a^{\frac{n-1}{2}}-1}{2}$$

$$k=2a^{\frac{n-1}{2}}$$

Demonstration:

$$a^n=\dfrac{(\dfrac{a^{\frac{n+1}{2}}-2a^{\frac{n-1}{2}}+1}{2})+{(\dfrac{a^{\frac{n+1}{2}}+2a^{\frac{n-1}{2}}-1}{2})}}{2}2a^{\frac{n-1}{2}}=$$

$$=\dfrac{2a^{\frac{n+1}{2}}}{4}2a^{\frac{n-1}{2}}=a^n$$

Examples:

1º) n=6, a=5

$$a_1=\dfrac{5^{\frac{6}{2}}+1}{2}=63$$

$$a_k=\dfrac{3*5^{\frac{6}{2}}-1}{2}=187$$

$$k=5^{\frac{6}{2}}=125$$

$$5^6=63+64+65+...+186+187=\dfrac{63+187}{2}125=15625$$

2º) n=7. a=4

$$a_1=\dfrac{4^{\frac{7+1}{2}}-2*4^{\frac{7-1}{2}}+1}{2}=64.5$$

$$a_k=\dfrac{4^{\frac{7+1}{2}}+2*4^{\frac{7-1}{2}}-1}{2}=191.5$$

$$k=2*4^{\frac{7-1}{2}}=128$$

$$4^7=64.5+65.5+66.5+...+190.5+191.5=\dfrac{64.5+191.5}{2}128=16384$$

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

Regards.