MHB Discover the Strange Pattern in Powers: A Question About Calculating Powers

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The discussion revolves around the observation of a pattern in the powers of integers, specifically how the differences between consecutive squares yield a constant second difference of 2. The user calculated the squares of numbers from 1 to 4 and noted the differences: 3, 5, and 7, leading to the second differences of 2. This pattern relates to the properties of quadratic functions, where a constant second difference indicates a quadratic relationship. The conversation highlights the connection between discrete sequences and their mathematical properties, emphasizing the significance of second differences in identifying quadratic functions. Understanding this concept is foundational in mathematics and calculus.
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Hello again,it has been a few minutes before my last thread and i am also pleased of its reply so thank about it,but i remembered another question i had,this time about powers.
One day i was calculating the powers of 1,2,3 until 10.This day i realized something strange.
1^2=1, 2^2=4, 3^2=9, 4^2=16...
so i did 4-1=3, 9-4=5 and 16-9=7...
and then i realized this:5-3=2 and 7-5=2... my question is why did it end up the number 2
so the next day i instantly asked my math teacher about it and he told me that this is already found but i am to young to understand why.So now i am asking help here (although i still think i won't understand since i think it will be complicated)

P.S I am sorry if you didn't understand the calculation but it was tough for me to explane it
Thank you
 
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1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
.
.
.

Do you see where the '2' is coming from now?
 
Angel1 said:
Hello again,it has been a few minutes before my last thread and i am also pleased of its reply so thank about it,but i remembered another question i had,this time about powers.
One day i was calculating the powers of 1,2,3 until 10.This day i realized something strange.
1^2=1, 2^2=4, 3^2=9, 4^2=16...
so i did 4-1=3, 9-4=5 and 16-9=7...
and then i realized this:5-3=2 and 7-5=2... my question is why did it end up the number 2
so the next day i instantly asked my math teacher about it and he told me that this is already found but i am to young to understand why.So now i am asking help here (although i still think i won't understand since i think it will be complicated)

P.S I am sorry if you didn't understand the calculation but it was tough for me to explane it
Thank you

$$(x+1)^2-x^2=2x+1$$

$$x^2-(x-1)^2=2x-1$$

$$2x+1-(2x-1)=2$$
 
Angel1 said:
Hello again,it has been a few minutes before my last thread and i am also pleased of its reply so thank about it,but i remembered another question i had,this time about powers.
One day i was calculating the powers of 1,2,3 until 10.This day i realized something strange.
1^2=1, 2^2=4, 3^2=9, 4^2=16...
so i did 4-1=3, 9-4=5 and 16-9=7...
and then i realized this:5-3=2 and 7-5=2... my question is why did it end up the number 2
so the next day i instantly asked my math teacher about it and he told me that this is already found but i am to young to understand why.So now i am asking help here (although i still think i won't understand since i think it will be complicated)

P.S I am sorry if you didn't understand the calculation but it was tough for me to explane it
Thank you

I like the way you are investigating how things work on your own. (Yes)

What you've discovered here is closely related to a result from the calculus. When the second derivative of a function is a constant, then the function will be quadratic. But, let's look at the discrete version. Suppose you are given the sequence:

5, 8, 14, 23, 35, 50, ...

And you are asked to find the $n$th term.

So, you look at the "first difference", that is, the difference between successive terms, and you find:

3, 6, 9, 12, 15

Then you look at the "second difference", that is the difference between successive terms of the first difference, and you find:

3, 3, 3, 3

So, we find a constant second difference, and we may state that the $n$th term of the sequence, which we'll call $a_n$, will be a quadratic in $n$:

$$a_n=An^2+Bn+C$$

To determine the unknown coefficients $(A,B,C)$, we may construct a system of equations based on the first 3 terms of the sequence, and their given values:

$$a_1=A+B+C=5$$

$$a_2=4A+2B+C=8$$

$$a_3=9A+3B+C=14$$

Solving this system, we obtain:

$$(A,B,C)=\left(\frac{3}{2},-\frac{3}{2},5\right)$$

And so we may state:

$$a_n=\frac{1}{2}(3n^2-3n+10)$$

What kind of general term do you suppose we'd get if we find the third difference is constant?
 
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