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Given sequence:

$A_1 = 0$ $A_2 = 1$ $A_3 = 3$ $A_4 = 6$ $A_5 = 10$ ...

And the recursive formula is given as $A_n = A_{n-1} + (n - 1)$

How can one find this sequence's explicit formula?

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Hint. One may write $$ A_n - A_{n-1} = (n - 1),\qquad n\ge2, $$ then summing from $2$ to $N$ terms telescope on the left hand side getting $$ A_N - A_1 = \sum_{n=2}^N(n - 1)=\frac{N(N-1)}2. $$

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We recognise $0,1,3,6,10,\cdots$ as $\displaystyle \left[\binom 12\right], \binom 22, \binom 32,\binom42, \binom 52,\cdots $ and conjecture that $$\color{red}{A_n=\binom n2}$$ Testing this in the recurrence relation given: $$A_n-A_{n-1}=\binom n2-\binom {n-1}2=\frac {n(n-1)}2-\frac {(n-1)(n-2)}2=n-1\\ \Longrightarrow A_n=A_{n-1}+(n-1)$$ Hence conjecture is true.

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Because $n-1=\frac{a_{n}-a_{n-1}}{n-(n-1)} \approx \frac{d}{dn} a_n$ is linear, it is reasonable to guess that $a_n$ is a quadratic because the integral of a linear function is a quadratic.

So guess $a_n=an^2+bn+c$.

This means,

$$\left(an^2+bn+c \right)-\left(a(n-1)^2+b(n-1)+c \right)$$

$$=a(2n-1)+b$$

$$=\color{red}{2a}n+\color{blue}{b-a}$$

$$:=\color{red}{1}n\color{blue}{-1}$$

We want:

$$2a=1$$

$$b-a=-1$$

$$a(1^2)+b(1)+c=0$$

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