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f(11) = (1+1)²
= 4
f²(11) = (f(11))
= f(4)
= 4² = 16
f³(11) = f(f²(11))
= f(16)
= (1+6)²
= 49
f^4 = f(f³(11))
= f(49)
= (4 + 9)²
= 169
f^5 = f(f^4))
= f(1 + 6 + 9)²
= 256
f^6 = f(f^5))
= f(256)
= (2 + 5 + 6)²
= 169
sampai pangkat 1998 terjadi pola berulang dari pangkat 4 sampai 1998
maka dirumuskan
f^1998 : f³
f^1995
karena pola berulang dua, maka dibagi 2
1995 : 2 = sisa 1
ambil titik pertama = 169