If f(a + b) = f(a) + f(b) - 2f(ab) for all nonnegative integers a and b, and f(1) = 1, compute f(1986)

f(1986) = 0

Step-by-step explanation:

f(a + b) = f(a) + f(b) - 2f(ab)

We need to find f(1986)

f(1986) = f(1 +1985). Using the above formula, we can write:

f(1 +1985) = f(1) + f(1985) - 2f(1 x 1985)

f(1986) = 1 - f(1985)                                         Equation 1

Applying the same formula again on f(1985), we get:

f(1985) = f(1 + 1984) = f(1) + f(1984) - 2f(1984)

f(1985) = 1 - f(1984)

Using this value in Equation 1, we get:

f(1986) = 1 - (1 - f(1984))

f(1986)= f(1984)

Continuing this, we can observe,

f(1986) = f(1984) = f(1982) = f(1980) = f(4) = f(2)

So,

f(1986) = f(2)

f(2) = f(1 + 1) = f(1) +f(1) - 2(1 x 1) = f(1) + f(1) - 2f(1)

f(2) = 1 + 1 - 2 = 0

Therefore,

f(1986) = f(2) = 0

i got 142 as my answer

step-by-step explanation:

$79500

step-by-step explanation: