Page 4 - FUNGSI PEMBANGKIT EKSPONENSIAL (FPE)
P. 4
Pembahasan Soal 1
0 1 2
Karena = 0 0! + 1 1! + 2 2! + ⋯ + ! dan = (0,0,1,1,1, … ), maka jika
kita subtitusikan:
0 0 3 3 3
= 0 → 0 = 0 = 0.1 = 0 = 3 → 3 = 1. =
0! 0! 3! 3! 3!
1 1 1 4 4 4
= 1 → 1 1! = 0 1! = 0. 1 = 0 = 4 → 4 4! = 1. 4! = 4! 4
2 2 2
= 2 → 2 2! = 1. 2! = 2!
Sehingga diperoleh:
2 3 4 2 3 4
= 0 + 0 + + + + ⋯ = + + + ⋯
2! 3! 4! 2! 3! 4!
