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

In summary, the conversation was about a study on numbers raised to an exponent and the discovery of a general formula to find the consecutive addends for odd and even numbers. Two cases were presented and examples were given to demonstrate the formulas. The speaker also suggested further testing and analysis, and cautioned against using the term "mente oscura" as the name for the theorem.
  • #1
mente oscura
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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.
 
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  • #2


Hello there,

Thank you for sharing your study on numbers raised to an exponent. It is always exciting to see patterns emerge in mathematics and to find general formulas that can explain them.

I did take a look at your examples and the formulas you have presented seem to hold true. However, I would suggest further testing and analysis to ensure that they hold true for all values of a, n, and k.

Also, I would like to point out that the term "mente oscura" translates to "dark mind" in English. I am not sure if that is the intended name for your theorem, but it is always important to choose names that accurately represent the concept being studied.

Overall, great work on your study and I encourage you to continue exploring and discovering patterns in mathematics.
 

FAQ: Express numbers, raised to an exponent, for sum of consecutive numbers.

What is an exponent?

An exponent is a mathematical notation that indicates the number of times a quantity is multiplied by itself. It is usually written as a superscript to the right of a number, such as 53, which means 5 multiplied by itself 3 times.

How do you express numbers raised to an exponent?

To express a number raised to an exponent, you simply write the base number followed by the exponent as a superscript. For example, to express 4 raised to the 3rd power, you would write 43.

What is the significance of sum of consecutive numbers?

The sum of consecutive numbers refers to the total when adding all numbers that come after each other in a sequence. This can be represented by the formula n(n+1)/2, where n is the number of consecutive numbers being added.

How do you use exponents to find the sum of consecutive numbers?

To find the sum of consecutive numbers using exponents, you can use the formula n(n+1)/2, where n is the number of consecutive numbers being added. For example, if you want to find the sum of the first 5 consecutive numbers (1+2+3+4+5), you would use the formula 5(5+1)/2 = 15.

Can you give an example of using exponents to find the sum of consecutive numbers?

Sure! Let's say we want to find the sum of the first 10 consecutive numbers (1+2+3+4+5+6+7+8+9+10). We can use the formula 10(10+1)/2 = 55, which tells us that the sum of the first 10 consecutive numbers is 55.

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