Page 266 - Buku Siap OSN Matematika SMP 2015(1)
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Solusi Olimpiade Matematika 2013
f(1) = 2000
f(x + 1) + 12 = f(x) f(x + 1) = f(x) – 12
Sehingga:
f(x + 1) = f(x) – 12
untuk x = 1
f(x + 1) = f(x) – 12
f(1 + 1) = f(1) – 12
f(2) = 2000 – 12
untuk x = 2
f(x + 1) = f(x) – 12
f(2 + 1) = f(2) – 12
f(3) = (2000 – 12) – 12
untuk x = 3
f(x + 1) = f(x) – 12
f(3 + 1) = f(3) – 12
f(4) = (2000 – 2(12)) – 12
untuk x = 4
f(x + 1) = f(x) – 12
f(4 + 1) = f(4) – 12
f(5) = (2000 – 3(12)) – 12
untuk x = x
f(x + 1) = f(x) – 12
f(x + 1) = [2000 – (x – 1)(12)] – 12
f(x + 1) = [2000 – (12x – 12)] – 12
f(x + 1) = 2000 – 12x + 12 – 12
Siap OSN Matematika SMP 2015 257
