Okay, so I have this problem where I need to find the sum of f(1) + f(2) + ... + f(1985), given that f(n + 1) = (-1)^{n + 1} * n - 2f(n) for n ≥ 1, and f(1) = f(1986). Hmm, let's see. I need to figure out a way to compute this sum. 

First, maybe I should try to find a pattern or a recurrence relation that can help me express f(n) in terms of previous terms. The given recurrence is f(n + 1) = (-1)^{n + 1} * n - 2f(n). Let me write that down again to make sure I have it right:

f(n + 1) = (-1)^{n + 1} * n - 2f(n)

So for each n starting at 1, the next term is this expression. Also, the problem tells me that f(1) = f(1986). That seems important. Since the recurrence relates f(n + 1) to f(n), and the function is periodic at the end (f(1) = f(1986)), maybe there's a way to use this periodicity to find the sum.

Let me consider writing out the first few terms to see if a pattern emerges. Let's start with n = 1:

For n = 1:
f(2) = (-1)^{2} * 1 - 2f(1) = 1 * 1 - 2f(1) = 1 - 2f(1)

For n = 2:
f(3) = (-1)^{3} * 2 - 2f(2) = -1 * 2 - 2f(2) = -2 - 2f(2)

But f(2) is 1 - 2f(1), so substituting that in:
f(3) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

For n = 3:
f(4) = (-1)^{4} * 3 - 2f(3) = 1 * 3 - 2f(3) = 3 - 2f(3)

Substituting f(3) from above:
f(4) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

Hmm, this is getting a bit messy. Let me see if there's a better approach. Maybe if I sum the recurrence relation over a range of n, that could help. Since the sum from f(1) to f(1985) is required, and the recurrence relates f(n + 1) to f(n), perhaps summing both sides from n = 1 to n = 1985 would help. Let's try that.

Sum from n = 1 to n = 1985 of f(n + 1) = sum from n = 1 to n = 1985 of [(-1)^{n + 1} * n - 2f(n)]

On the left side, the sum becomes sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k), since when n=1, k=2, and when n=1985, k=1986.

On the right side, we have sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

So putting it all together:

sum_{k=2}^{1986} f(k) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

But the left side is sum_{k=2}^{1986} f(k) = sum_{k=1}^{1986} f(k) - f(1)

Similarly, the right side is sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

Therefore, substituting back:

sum_{k=1}^{1986} f(k) - f(1) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

Let me denote S = sum_{n=1}^{1985} f(n). Then, sum_{k=1}^{1986} f(k) = S + f(1986). But given that f(1) = f(1986), we can write sum_{k=1}^{1986} f(k) = S + f(1). Therefore, substituting into the equation:

S + f(1) - f(1) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2S

Simplify left side: S + f(1) - f(1) = S

So we have:

S = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2S

Bring the 2S to the left:

S + 2S = sum_{n=1}^{1985} [(-1)^{n + 1} * n]

3S = sum_{n=1}^{1985} [(-1)^{n + 1} * n]

Thus, S = (1/3) * sum_{n=1}^{1985} [(-1)^{n + 1} * n]

So if I can compute that sum, then S is one-third of that. Therefore, the problem reduces to computing the sum of (-1)^{n + 1} * n from n=1 to 1985, then dividing by 3.

Let me compute the sum sum_{n=1}^{1985} [(-1)^{n + 1} * n]. Let's denote this sum as T.

So T = sum_{n=1}^{1985} [(-1)^{n + 1} * n]

Let me write out the terms:

For n=1: (-1)^{2} * 1 = 1*1 = 1

n=2: (-1)^3 * 2 = -1*2 = -2

n=3: (-1)^4 * 3 = 1*3 = 3

n=4: (-1)^5 *4 = -1*4 = -4

... and so on.

So the sequence is 1, -2, 3, -4, 5, -6, ..., up to n=1985.

Since 1985 is odd, the last term is positive. Let me group the terms in pairs:

(1 - 2) + (3 - 4) + (5 - 6) + ... + (1983 - 1984) + 1985

Each pair (odd - even) is (-1). How many such pairs are there?

From n=1 to n=1984, that's 1984 terms, which is 992 pairs. Each pair sums to -1, so total sum for pairs: 992*(-1) = -992

Then add the last term, which is 1985.

Therefore, T = -992 + 1985 = 1985 - 992 = 993

Therefore, T = 993

Therefore, S = T / 3 = 993 / 3 = 331

So the sum S = 331. Therefore, the answer is 331. But let me verify this calculation again to be sure.

First, when n is odd: (-1)^{n + 1} = (-1)^{even} = 1, so term is positive.

When n is even: (-1)^{n + 1} = (-1)^{odd} = -1, so term is negative.

Therefore, the series is 1 - 2 + 3 - 4 + 5 - 6 + ... + 1985

Yes, that's correct.

Grouping into pairs: (1 - 2) + (3 - 4) + ... + (1983 - 1984) + 1985

Each pair is -1, and there are 1984 terms before the last term, so 1984 / 2 = 992 pairs.

Each pair is -1, so total from pairs: 992*(-1) = -992

Then add the last term, 1985: -992 + 1985 = 1985 - 992 = 993. Correct.

Then S = 993 / 3 = 331. So the answer is 331. But let me check once more.

Alternatively, maybe there's a formula for the sum T. Let's recall that for an alternating sum of integers, the formula can be derived.

Suppose we have T = sum_{k=1}^{m} (-1)^{k + 1} * k

If m is even: sum = -m/2

If m is odd: sum = (m + 1)/2

Wait, let's check for small m.

For m=1: 1. Formula: (1 + 1)/2 = 1. Correct.

m=2: 1 - 2 = -1. Formula: -2/2 = -1. Correct.

m=3: 1 - 2 + 3 = 2. Formula: (3 + 1)/2 = 2. Correct.

m=4: 1 - 2 + 3 - 4 = -2. Formula: -4/2 = -2. Correct.

So the formula is:

If m is even: T = -m / 2

If m is odd: T = (m + 1)/2

In our case, m = 1985, which is odd.

Therefore, T = (1985 + 1)/2 = 1986 / 2 = 993. Which matches our previous calculation. Therefore, S = 993 / 3 = 331. So answer is 331.

Wait, but the problem says f(1986) = f(1). So the sum from n=1 to 1985 is S = 331. But let me verify if this is correct.

Alternatively, maybe there's a mistake in the initial step when we set up the equation. Let's go through that again.

We had:

sum_{n=1}^{1985} f(n + 1) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

Left side: sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k) = sum_{k=1}^{1986} f(k) - f(1)

Right side: sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

Therefore:

sum_{k=1}^{1986} f(k) - f(1) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2 * sum_{n=1}^{1985} f(n)

But sum_{k=1}^{1986} f(k) = sum_{n=1}^{1985} f(n) + f(1986) = S + f(1986)

But since f(1986) = f(1), we have sum_{k=1}^{1986} f(k) = S + f(1)

Therefore, substituting:

S + f(1) - f(1) = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2S

Which simplifies to S = sum_{n=1}^{1985} [(-1)^{n + 1} * n] - 2S

Then, 3S = sum ... so S = sum / 3. Which gives S = 993 / 3 = 331.

Therefore, the answer is 331. So I think this is correct.

But let me check again with a smaller example to make sure the approach is right.

Suppose instead of 1985, the upper limit is 3. Let's say n=1 to 3, and f(4)=f(1). Let's compute the sum S = f(1) + f(2) + f(3).

Given the recurrence:

f(2) = (-1)^2 * 1 - 2f(1) = 1 - 2f(1)

f(3) = (-1)^3 * 2 - 2f(2) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

f(4) = (-1)^4 * 3 - 2f(3) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

But f(4) = f(1), so:

11 - 8f(1) = f(1)

11 = 9f(1)

f(1) = 11/9

Then, S = f(1) + f(2) + f(3) = 11/9 + (1 - 2*(11/9)) + (-4 + 4*(11/9))

Compute each term:

f(1) = 11/9

f(2) = 1 - 22/9 = (9 - 22)/9 = -13/9

f(3) = -4 + 44/9 = (-36 + 44)/9 = 8/9

Sum S = 11/9 -13/9 +8/9 = (11 -13 +8)/9 = 6/9 = 2/3

Now, according to our formula:

sum_{n=1}^{3} [(-1)^{n + 1} * n] = 1 - 2 + 3 = 2

Then, S = 2 / 3 = 2/3. Which matches the manual calculation. Therefore, the formula works for this case.

Therefore, the method seems correct. So in the original problem, the answer is 331. Therefore, the final answer is \boxed{331}

**Final Answer**
\boxed{331}
Therefore, the value of the sum is $\boxed{331}$.