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Question Number 32985 by soufiane zarik last updated on 08/Apr/18

	Answered by MJS last updated on 08/Apr/18

	$a_{1} = 1$ $a_{2} = a_{1 + 1} = 2 a_{1} + 1 = 3$ $a_{3} = a_{1 + 2} = a_{1} + a_{2} + 2 = 6$ $a_{4} = a_{1 + 3} = a_{1} + a_{3} + 3 = 10$ $(a_{4} = a_{2 + 2} = 2 a_{2} + 4 = 10)$ $a_{5} = a_{1 + 4} = a_{1} + a_{4} + 4 = 15$ $(a_{5} = a_{2} + a_{3} = a_{2} + a_{3} + 6 = 15)$ $a_{n} = a_{1} + a_{n - 1} + 1 \times (n - 1) = a_{n - 1} + n$ $a_{n} = \underset{i = 1}{\sum^{n}} i = \frac{n}{2} (n + 1)$ $proof :$ $a_{m + n} = \frac{m + n}{2} (m + n + 1) =$ $= \frac{m^{2} + 2 m n + n^{2} + m + n}{2} =$ $= \frac{m^{2} + m}{2} + \frac{n^{2} + n}{2} + m n =$ $= \frac{m}{2} (m + 1) + \frac{n}{2} (n + 1) + m n =$ $= a_{m} + a_{n} + m n$ $a_{100} = 5050$
	Commented by soufiane zarik last updated on 08/Apr/18

	$thank you very much sir!$
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