Question

If $f(x+1)=(-1)^{n+1}x-2f(x)$ for $x \in N$ and $f(1)=f(1986)$, Then sum of digits of $(f(1)+f(2)+...+f(1985))$ is

• A. 26
• B. 7
• C. 18
• D. 20

Answer 1

Answer: (A)

26

Answer 2

Let the given functional equation be $f(x+1) = C \cdot x - 2f(x)$, where $C = (-1)^{n+1}$ is a constant. We are given $f(1) = f(1986)$. We need to find the sum of digits of $S = \sum_{x=1}^{1985} f(x)$.

Summing the functional equation from $x=1$ to $1985$: $\sum_{x=1}^{1985} f(x+1) + 2 \sum_{x=1}^{1985} f(x) = \sum_{x=1}^{1985} C \cdot x$ The first term is $\sum_{x=1}^{1985} f(x+1) = f(2) + f(3) + \dots + f(1986)$. This can be written as $(f(1) + f(2) + \dots + f(1985) + f(1986)) - f(1) = S + f(1986) - f(1)$. The equation becomes: $(S + f(1986) - f(1)) + 2S = C \sum_{x=1}^{1985} x$ Given $f(1) = f(1986)$, the term $f(1986) - f(1) = 0$. So, the equation simplifies to: $3S = C \sum_{x=1}^{1985} x$ The sum of the first 1985 natural numbers is $\frac{1985 \cdot (1985+1)}{2} = \frac{1985 \cdot 1986}{2} = 1985 \cdot 993$. $3S = C \cdot (1985 \cdot 993)$ $S = C \cdot \frac{1985 \cdot 993}{3}$ $S = C \cdot (1985 \cdot 331)$ Calculating $1985 \cdot 331$: $1985 \times 331 = 657035$. So, $S = C \cdot 657035$.

Since $C = (-1)^{n+1}$, $C$ can be either $1$ or $-1$. If $C = 1$, then $S = 657035$. The sum of digits is $6+5+7+0+3+5 = 26$. If $C = -1$, then $S = -657035$. The sum of digits of a negative number is usually taken for its absolute value. The sum of digits of $|S| = 657035$ is $6+5+7+0+3+5 = 26$.

In either case, the sum of digits is 26.

The final answer is $\boxed{26}$.