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

[Angel11]

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

 

[Greg]

1
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?

 

[Fermat1]

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

\(\displaystyle (x+1)^2-x^2=2x+1\)

\(\displaystyle x^2-(x-1)^2=2x-1\)

\(\displaystyle 2x+1-(2x-1)=2\)

 

[MarkFL]

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 $n$:

\(\displaystyle a_n=An^2+Bn+C\)

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:

\(\displaystyle a_1=A+B+C=5\)

\(\displaystyle a_2=4A+2B+C=8\)

\(\displaystyle a_3=9A+3B+C=14\)

Solving this system, we obtain:

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

And so we may state:

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

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