We start with f(n) = - 2f(n-1) + 1 and plug in n = 2 which yields
f(2) = - 2f(1) + 1
It is given that f(1) = 3. Therefore, we have
f(2) = - 2(3) + 1 = - 6 + 1 = -5
Now that we have f(2), we can plug in n = 3 in the given equation:
f(3) = - 2f(2) + 1 = - 2(-5) + 1 = 10 + 1 = 11
Going on, we plug in n = 4 in the given equation:
f(4) = - 2f(3) + 1 = - 2(11) + 1 = - 22 + 1 = -21
Finally, we plug in n = 5 in the given equation:
f(5) = - 2f(4) + 1 = - 2(-21) + 1 = 42 + 1 = 43
So, f(5) = 43
METHOD 2:
Plug in n = 5 in the given equation to get
f(5) = -2f(4) + 1
f(5) = -2(-2f(3) + 1) + 1
f(5) = 4f(3) - 1
f(5) = 4(-2f(2) + 1) - 1
f(5) = -8f(2) + 3
f(5) = -8(-2f(1) + 1) + 3
f(5) = 16f(1) - 5
f(5) = 16(3) - 5
f(5) = 48 - 5
f(5) = 43
