Discussion
If \(f: R \rightarrow R\) satisfies the relation \(f(x + y) = f(x) + f(y), \forall x, y \in R\) and \(f(1) = 3\) then \(f(0) + f(1) + f(2) + f(3)\) is equal to
Answer: D
The sum is 18.
A function with \(f(x+y) = f(x) + f(y)\) is additive. Putting \(x = y = 0\) gives \(f(0) = 2f(0)\), so \(f(0) = 0\). Repeated use gives \(f(n) = n f(1)\) for whole numbers, so with \(f(1) = 3\):
\[ \begin{aligned} f(2) &= f(1) + f(1) = 6 \\ f(3) &= f(2) + f(1) = 9 \\ \text{sum} &= 0 + 3 + 6 + 9 = 18 \end{aligned} \]
Once \(f(0) = 0\) and \(f(n) = 3n\) are fixed, the sum is determined, so 12, 14, 16 and 22 cannot occur. The common slip is to assume \(f(0) = f(1)\), which gives 21, not an option; 12 is \(f(1) + f(3)\) alone.
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