Question-32985

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!