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It supposed to be some thing simple, but some how I don't get it.

I have a function with constant rate increase of 2. How can I calculate the n'th number?

It starts with 1, and it looks like this:

$$1,3,7,13,21,31,43...$$

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2 Answers

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$f(n)=n^2$ goes $1,4,9,16,25,36,49$, it has the correct rate increase but the increase starts too fast.

$f(n)=n^2-n$ goes $0,2,6,12,20,30,42$ has the correct rate increase and initial increase. but it starts too small.

$f(n)=n^2-n+1$ fits perfectly.

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First observe that the second differences are constant.

Then we can assume it is $$f(x) = ax^2 + bx + c.$$

Subbing in $x = 0,1,2$ we have the system of equations

$$\pmatrix{1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 3 \\ 1 & 2 & 4 & 7} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 2 & 4 & 6} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 2 & 2} = \pmatrix{1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1}.$$

Thus, we have $f(x) = x^2 + x + 1$.

For your interests.

If $$a = 1 + 2 + \ldots + n, $$ then it is also true that $$2a = (1 + 2 + \ldots + n) + (n + (n - 1) + \ldots + 1) = (n + 1) + \ldots (n + 1) = n(n +1).$$

Thus,

$$ 1 + 2 + \ldots + n = \frac{n(n+1)}{2}.$$

Finally, if $T(0) = 1$, then your sequence is just

$$T(n) = T(0) + 2 + 4 + \ldots + 2n = 1 + 2(1 + 2 + \ldots + n) = 1 + 2\frac{n(n+1)}{2} = n^2 + n + 1.$$

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