Is there a pattern between the sequence of cubes and the derivative of x^3?

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The discussion explores the relationship between the derivative of the function x^2, which is 2x, and the differences in the sequence of perfect squares (1, 4, 9, 16, ...), where the differences are 3, 5, 7, 9, etc. It highlights that the difference between consecutive squares can be expressed as 2x + 1, illustrating why the differences are consistently odd numbers. Additionally, the conversation suggests examining the sequence of cubes (1, 8, 27, 64, ...) to identify potential patterns in their differences. The inquiry emphasizes the mathematical connections between derivatives and sequences. Understanding these relationships can deepen insights into calculus and algebraic patterns.
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Is there a relation between the fact that the derivative of x^2 is 2x and that the difference between 1,4,9,16, ... is 3, 5, 7, 9, ...?

And why is the difference always 2?
 
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##(x+1)²-x²=2x+1##

This shows why the difference is always 2.
 
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jaja1990 said:
Is there a relation between the fact that the derivative of x^2 is 2x and that the difference between 1,4,9,16, ... is 3, 5, 7, 9, ...?

And why is the difference always 2?

If you spotted that you should look for a pattern in the sequence ##x^3: 1, 8, 27, 64 \dots##
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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