Question

Let a,b $\epsilon$ R. If f(x)= ax+bis such that
a+b=4 and f(x + y) = f(x)+f(y)-2 for all x, y $\epsilon$ R,
then  $ \sum_{n=1} ^{50} f(n)$ =__________  (in integer).

Answer 1

Correct Answer: 2650

Given the problem, we need to determine the value of $ \sum_{n=1} ^{50} f(n)$ where $f(x) = ax + b$. We know that $a+b = 4$ and $f(x + y) = f(x) + f(y) - 2$ for all $x, y \in \mathbb{R}$. Let's solve this step by step:

1. First, the condition $f(x+y) = f(x) + f(y) - 2$ can be expanded as: $$a(x+y) + b = ax + b + ay + b - 2.$$ Simplifying gives us $$ax + ay + b = ax + ay + 2b - 2.$$ This means that $2b - 2 = b$, leading to $b = 2$.
2. Substituting $b = 2$ into $a + b = 4$ yields $a = 2$.
3. Thus the function becomes $f(x) = 2x + 2$.
4. Now, compute the required sum $ \sum_{n=1} ^{50} f(n) = \sum_{n=1} ^{50} (2n + 2) = 2\sum_{n=1} ^{50} n + \sum_{n=1} ^{50} 2$.
5. The first part is $2\sum_{n=1} ^{50} n = 2\frac{50 \times 51}{2} = 2550.$
6. The second part is simply $2 \times 50 = 100.$
7. Hence, $ \sum_{n=1} ^{50} f(n) = 2550 + 100 = 2650.$

Therefore, the final result is confirmed as 2650.