If f(n) = f(n - 1) + 4, then
f(n - 1) = f(n - 2) + 4
and substituting this into the first equation gives
f(n) = f(n - 2) + 2•4
Similarly,
f(n - 2) = f(n - 3) + 4
so that
f(n) = f(n - 3) + 3•4
and so on.
Note the pattern:
f(n) = f(n - k) + k•4
so that when k = n - 1, we have
f(n) = f(n - (n - 1)) + (n - 1)•4
or
f(n) = f(1) + 4 (n - 1)
so that
[D] f(n) = -2 + 4 (n - 1)