<Question>
If
\[f(n + 1) = (-1)^{n + 1} n - 2f(n)\]for $n \ge 1,$ and $f(1) = f(1986),$ compute
\[f(1) + f(2) + f(3) + \dots + f(1985).\]
</Question>

<think>
<Plan_1>
Okay, so I need to solve this recurrence relation problem. The function f is defined for integers n ≥ 1 with the recurrence relation f(n + 1) = (-1)^(n + 1) * n - 2f(n), and it's given that f(1) = f(1986). The goal is to compute the sum f(1) + f(2) + ... + f(1985). Hmm, that's quite a range. Let me try to break this down step by step.

First, let's understand the recurrence relation. For each n ≥ 1, the next term f(n + 1) is equal to (-1)^(n + 1) multiplied by n minus twice the current term f(n). So each term depends on the previous term, with an alternating sign coefficient and a subtraction of twice the previous term. The initial condition is that f(1) equals f(1986), which suggests some periodicity or cyclicity in the function over 1985 steps.

Since the problem involves a recurrence relation and a condition that links the first and the 1986th term, perhaps there's a pattern that repeats every certain number of terms? Or maybe when we sum all the terms from f(1) to f(1985), the recurrence allows us to express the sum in a simpler form?

Let me denote the sum S = f(1) + f(2) + ... + f(1985). If I can find a relationship involving S, maybe through manipulating the recurrence relation, then perhaps I can solve for S.

Let's try to express each f(n + 1) in terms of f(n) and substitute into the sum. Wait, but the sum is from f(1) to f(1985), and the recurrence relates f(n + 1) to f(n). So if I can write f(2) in terms of f(1), f(3) in terms of f(2), and so on, maybe I can express the entire sum in terms of f(1), and then use the condition f(1) = f(1986) to find f(1), and subsequently compute the sum.

Let me try that approach.

<prompt_1.1> Express \( f(2) \) in terms of \( f(1) \) using the recurrence relation and the given condition \( f(1) = f(1986) \). </prompt_1.1>
-----

But let me check with a smaller case to see if this makes sense. Let's take a smaller N instead of 1985, say N=3, so that we have f(1)=f(5). Let's see if the same logic applies.

<prompt_1.2> Test the logic with a smaller case where \( N = 3 \) and \( f(1) = f(5) \). </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Starting with n = 1:

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

Similarly, for n = 2:

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

But since we already expressed f(2) in terms of f(1), we can substitute that:

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

Continuing to n = 3:

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

For n = 4:

f(5) = (-1)^(4 + 1) * 4 - 2f(4) = (-1)^5 * 4 - 2*(11 - 8f(1)) = -4 - 22 + 16f(1) = -26 + 16f(1).

But we know that f(5) = f(1). So setting f(5) equal to f(1):

f(1) = -26 + 16f(1).

Solving for f(1):

f(1) - 16f(1) = -26 => -15f(1) = -26 => f(1) = 26/15.
</execution_1.1>

<execution_1.2>
Wait, maybe N=4? Let's take N=4, so f(1)=f(5). Let's compute the sum S = f(1)+f(2)+f(3)+f(4).

First, write the recurrence relations:

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

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

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

For n=4: f(5) = (-1)^(4+1)*4 - 2f(4) = (-1)^5*4 - 2f(4) = -4 - 2f(4)

But since f(5) = f(1), so:

f(1) = -4 - 2f(4)

But f(4) is expressed in terms of f(3), which is in terms of f(2), which is in terms of f(1). Let's substitute backwards.

First, express f(4):

f(4) = 3 - 2f(3) = 3 - 2*(-2 - 2f(2)) = 3 + 4 + 4f(2) = 7 + 4f(2)

But f(2) = 1 - 2f(1), so:

f(4) = 7 + 4*(1 - 2f(1)) = 7 + 4 - 8f(1) = 11 - 8f(1)

Then, f(5) = -4 - 2*(11 - 8f(1)) = -4 -22 + 16f(1) = -26 + 16f(1)

But f(5) = f(1), so:

f(1) = -26 + 16f(1)

Subtract f(1) from both sides:

0 = -26 + 15f(1)

Thus, 15f(1) = 26 => f(1) = 26/15

Then, compute the sum S = f(1) + f(2) + f(3) + f(4):

First, f(1) = 26/15

f(2) = 1 - 2*(26/15) = 1 - 52/15 = (15 - 52)/15 = (-37)/15

f(3) = -2 - 2*(-37/15) = -2 + 74/15 = (-30/15 + 74/15) = 44/15

f(4) = 11 - 8*(26/15) = 11 - 208/15 = (165/15 - 208/15) = (-43)/15

Sum S = 26/15 - 37/15 + 44/15 -43/15 = (26 -37 +44 -43)/15 = (-10 +44 -43)/15 = (34 -43)/15 = (-9)/15 = -3/5

But according to the formula S = (-1)^{N+1}/3 * (f(1) + f(N+1)). Here, N=4, so S = (-1)^5 /3 * (f(1) + f(5)) = (-1)/3*(26/15 +26/15) = (-1)/3*(52/15) = -52/45 ≈ -1.155... But our manual calculation gave -3/5 = -0.6. These are different. Hmm, so that suggests that the formula might not hold for smaller N? Or perhaps I made a miscalculation here.

Wait, maybe my initial assumption about the formula is incorrect for smaller N. Let me check again.

Wait, in the original problem, N=1985, which is one less than 1986. So in the small case, N=4, the sum is up to f(4), and f(5)=f(1). But when I computed manually, the sum was -3/5, but according to the supposed formula, it would be -52/45. These are different. So perhaps the formula S = (-1)^{N+1}/3*(f(1)+f(N+1)) is not generally valid, or maybe I made a mistake in deriving it.

Alternatively, maybe there's a different approach. Let's see.

Alternatively, perhaps consider writing the recurrence as f(n+1) + 2f(n) = (-1)^{n+1} n. Then, this is a linear nonhomogeneous recurrence relation. Maybe we can solve it using standard techniques.

The homogeneous recurrence would be f(n+1) + 2f(n) = 0. The characteristic equation is r + 2 = 0, so r = -2. Therefore, the general solution to the homogeneous equation is f_h(n) = A*(-2)^n, where A is a constant.

Then, we need a particular solution to the nonhomogeneous equation. The nonhomogeneous term is (-1)^{n+1} n, which is a product of (-1)^n and n. Since the nonhomogeneous term is of the form P(n)*(-1)^n, where P(n) is a polynomial of degree 1, we can try a particular solution of the form f_p(n) = (Bn + C)(-1)^n.

Let's substitute f_p(n) into the recurrence equation:

f_p(n+1) + 2f_p(n) = (-1)^{n+1} n

Compute f_p(n+1):

f_p(n+1) = [B(n+1) + C](-1)^{n+1}

Then, f_p(n+1) + 2f_p(n) = [B(n+1) + C](-1)^{n+1} + 2[Bn + C](-1)^n

Factor out (-1)^n:

= [B(n+1) + C](-1)*(-1)^n + 2[Bn + C](-1)^n

= [ -B(n+1) - C + 2Bn + 2C ] (-1)^n

Simplify the expression inside the brackets:

- Bn - B - C + 2Bn + 2C = ( -Bn + 2Bn ) + ( -B ) + ( -C + 2C )

= Bn + (-B) + C

So, we have:

[Bn + (-B + C)] (-1)^n = (-1)^{n+1} n = (-1)^n (-1) n

Therefore, equate the coefficients:

For the term involving n: Bn must equal -1*n. Therefore, B = -1.

For the constant term: (-B + C) must equal -1 (since the right-hand side is (-1)^n*(-1)n, which has coefficient -1 for the constant term when n is 0? Wait, actually, the right-hand side is (-1)^n*(-1)n, so the coefficient for n is -1, and there is no constant term. Wait, but in our expression, the left-hand side after substitution has a term with n and a constant term. Therefore, to match the right-hand side, which is (-1)^n*(-1)n, we need:

Coefficient of n: B = -1

Constant term: (-B + C) = 0, since there is no constant term on the right-hand side.

Given B = -1, then:

-(-1) + C = 0 => 1 + C = 0 => C = -1

Therefore, the particular solution is f_p(n) = (-n -1)(-1)^n = (n + 1)(-1)^{n + 1}

Therefore, the general solution to the recurrence is:

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

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

Now, apply the initial condition f(1) = f(1986). Let's compute f(1) and f(1986) using this general solution.

First, compute f(1):

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

Compute f(1986):

f(1986) = A*(-2)^{1986} + (1986 + 1)(-1)^{1987} = A*(2)^{1986} + 1987*(-1)^{1987} = A*2^{1986} - 1987

But since f(1) = f(1986), set them equal:

-2A + 2 = A*2^{1986} - 1987

Solve for A:

-2A - A*2^{1986} = -1987 - 2

A*(-2 - 2^{1986}) = -1989

Multiply both sides by -1:

A*(2 + 2^{1986}) = 1989

Therefore:

A = 1989 / (2 + 2^{1986}) = 1989 / [2(1 + 2^{1985})] = (1989 / 2) / (1 + 2^{1985})

Hmm, that's a very small number since 2^{1985} is enormous. So A is approximately (1989/2) * (1 / 2^{1985}) ) = 1989 / 2^{1986}, which is extremely close to zero.
-----

But let's compute the sum S = Σ_{n=1}^{1985} f(n). Let's substitute the general solution into the sum.


-----

Wait, but let's check this with the smaller case where N=4. If we use the formula S = (-1)^{N+1}/3 * (f(1) + f(N+1)), then for N=4, S = (-1)^5 /3 * (f(1) + f(5)) = (-1)/3*(26/15 +26/15) = (-52)/45 ≈ -1.155...., but our manual calculation gave -3/5 = -0.6. These don't match. Therefore, the formula might not be generalizable. Hmm, so maybe my initial approach was incorrect.

Alternatively, perhaps the formula is correct, but in the smaller case, there was an error in calculation. Let's check again.

Wait, in the smaller case with N=4, we found f(1)=26/15, and the sum S=-3/5. Let's compute (-1)^{5}/3*(f(1)+f(5))= (-1)/3*(26/15 +26/15)= (-1)/3*(52/15)= -52/45≈-1.155. But the actual sum is -3/5=-0.6. These are different. Therefore, the formula S = (-1)^{N+1}/3*(f(1)+f(N+1)) is not valid. So my previous assumption was wrong.

Therefore, we need another approach. Let's go back to the general solution.

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

We need to compute the sum S = Σ_{n=1}^{1985} f(n) = Σ_{n=1}^{1985} [A*(-2)^n + (n + 1)(-1)^{n + 1} ] = A*Σ_{n=1}^{1985} (-2)^n + Σ_{n=1}^{1985} (n + 1)(-1)^{n + 1}

Let's compute each sum separately.

First, compute the sum Σ_{n=1}^{1985} (-2)^n. This is a geometric series. The sum of a geometric series Σ_{n=1}^N r^n = r*(1 - r^N)/(1 - r). Here, r = -2.

So, Σ_{n=1}^{1985} (-2)^n = (-2)*(1 - (-2)^{1985})/(1 - (-2)) = (-2)*(1 - (-2)^{1985})/3 = [ -2 + 2*(-2)^{1985} ] / 3

Similarly, compute the second sum: Σ_{n=1}^{1985} (n + 1)(-1)^{n + 1}

Let me adjust the index: let k = n + 1, so when n=1, k=2, and when n=1985, k=1986. So the sum becomes Σ_{k=2}^{1986} k*(-1)^{k}

But Σ_{k=2}^{1986} k*(-1)^k = Σ_{k=1}^{1986} k*(-1)^k - 1*(-1)^1 = Σ_{k=1}^{1986} k*(-1)^k + 1

Therefore, we need to compute Σ_{k=1}^{1986} k*(-1)^k + 1

The sum Σ_{k=1}^{N} k*(-1)^k is a known series. Let's recall that:

Σ_{k=1}^{N} k*(-1)^k = (-1)^N * (N + 1)/2 if N is even, and (-1)^N * N/2 if N is odd. Wait, let me verify.

Let's compute for small N:

N=1: -1 → formula: (-1)^1*(1 + 1)/2 = -1*2/2 = -1 ✔️

N=2: -1 + 2*(-1)^2 = -1 + 2 = 1. Formula: (-1)^2*(2 + 1)/2 = 1*3/2 = 1.5 ❌ Doesn't match. Hmm.

Wait, perhaps another approach. Let's consider the sum S = Σ_{k=1}^{N} k*(-1)^k

We can write S = -1 + 2 -3 +4 -5 +... ±N

This can be grouped as:

(-1 +2) + (-3 +4) + ... + [up to N terms]

Each pair is 1, and the number of pairs is N/2 if N even, or (N-1)/2 if N odd, with a remaining term if N is odd.

Therefore, if N is even:

S = (N/2)*1 = N/2

If N is odd:

S = [(N -1)/2]*1 + (-N) = (N -1)/2 - N = (-N -1)/2

But let's check:

For N=1: (-1 -1)/2 = -1 ✔️

For N=2: 2/2 =1 ✔️

For N=3: (-3 -1)/2 = -2. Let's compute manually: -1 +2 -3 = -2 ✔️

For N=4: 4/2 =2. Manually: -1 +2 -3 +4 = 2 ✔️

Therefore, the formula is:

Σ_{k=1}^{N} k*(-1)^k = { N/2, if N even; (-N -1)/2, if N odd }

Therefore, in our case, N=1986, which is even. Therefore:

Σ_{k=1}^{1986} k*(-1)^k = 1986/2 = 993

Therefore, Σ_{k=2}^{1986} k*(-1)^k = 993 + 1 = 994

Wait, wait. Wait, original sum was Σ_{k=2}^{1986} k*(-1)^k = Σ_{k=1}^{1986} k*(-1)^k - 1*(-1)^1 = 993 - (-1) = 993 +1 = 994. Correct.

Therefore, the second sum is 994.

Putting it all together:

S = A*[ (-2 + 2*(-2)^{1985} ) /3 ] + 994

But recall that A = 1989 / (2 + 2^{1986})

So, substitute A:

S = [1989 / (2 + 2^{1986})] * [ (-2 + 2*(-2)^{1985} ) /3 ] + 994

Simplify the expression inside the brackets:

First, compute 2*(-2)^{1985}:

Note that (-2)^{1985} = (-1)^{1985} * 2^{1985} = -2^{1985} (since 1985 is odd)

Therefore, 2*(-2)^{1985} = 2*(-2^{1985}) = -2^{1986}

Therefore, the numerator becomes:

-2 + (-2^{1986}) = -2 - 2^{1986} = - (2 + 2^{1986})

Therefore, the first term becomes:

[1989 / (2 + 2^{1986})] * [ - (2 + 2^{1986}) /3 ] = 1989 * (-1)/3 = -1989 /3 = -663

Therefore, S = -663 + 994 = 331

Wait, 994 - 663 = 331. So the sum S is 331.
-----

Therefore, the sum f(1) + f(2) + ... + f(1985) is \boxed{\dfrac{331}{1}}.

**Final Answer**
\boxed{331}
-----

**Final Answer**
\boxed{331}
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1:
But let's check this with the smaller case where N=4. If we use the formula S = (-1)^{N+1}/3 * (f(1) + f(N+1)), then for N=4, S = (-1)^5 /3 * (f(1) + f(5)) = (-1)/3*(26/15 +26/15) = (-52)/45 ≈-1.155...., but our manual calculation gave -3/5 = -0.6. These are different. Therefore, the formula might not be generalizable.

Alternatively, perhaps the formula S = (-1)^{N+1}/3 * (f(1) + f(N+1)) is not valid. Let's verify with the original problem.

In the original problem, N=1985, so the sum is up to f(1985). Then f(1986) = f(1). Therefore, using the general solution, we can compute S.

But given the complexity of the expression, and that we arrived at a concrete answer of 331 through the general solution approach, which seems consistent, I will go with that.
-----

**Final Answer**
\boxed{331}
</Plan_2>

<Execution_2>
No prompt to execute.
</Execution_2>

<Final_answer>
Therefore, the sum of the sequence from \(f(1)\) to \(f(1985)\) is \(\boxed{331}\).
</Final_answer>
</think>