We are given the two equations:
H(n) = 1 if n = 1 --> eqtn 1
H(n) = H(n-1) + 6n – 6 if n>1 --> eqtn 2
Since we are to find for H(9), so obviously we would use eqtn 2. However take note that in eqtn 2, we have a factor which says H(n-1), this means that we have to calculate also for the value of H(8). However for H(8) we also have to calculate for H(7), therefore we need to calculate all values of H from 8 to 1 also. Let us start from 1:
H(1) = 1
H(2) = 1 + 6(2) – 6 = 7
H(3) = 7 + 6(3) – 6 = 19
H(4) = 19 + 6(4) – 6 = 37
H(5) = 37 + 6(5) – 6 = 61
H(6) = 61 + 6(6) – 6 = 91
H(7) = 91 + 6(7) – 6 = 127
H(8) = 127 + 6(8) – 6 = 169
H(9) = 169 + 6(9) – 6 = 217
Therefore the answer is:
H(9) = 217