If f: R -> R satisfies f(x+y)=f(x)+f(y), for all x, y in R and f(1)=7, then the sum from r=1 to n of f(r) is: A 7n(n+1)/2 B 7n/2 C 7(n+1)/2 D 7n + (n+1)
Question
If $f: R \rightarrow R$ satisfies $f(x+y)=f(x)+f(y)$, for all $x, y \in R$ and $f(1)=7$, then $\sum_{r=1}^{n} f(r)$ is:
A $\frac{7 n(n+1)}{2}$ B $\frac{7 n}{2}$ C $\frac{7(n+1)}{2}$ D $7n + (n+1)$
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Answer
:
Given that the function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the property
$$ f(x+y) = f(x) + f(y) \quad \forall x, y \in \mathbb{R}, $$
and also given that $f(1) = 7$, we need to determine the value of the summation $\sum_{r=1}^{n} f(r)$.
First, let's analyze the given functional equation. The equation
$$ f(x+y) = f(x) + f(y) $$
is a well-known Cauchy functional equation. One of the standard solutions to this equation is a linear function, specifically, $f(x) = mx$ for some constant $m$.
Given
$$ f(1) = 7, $$
we substitute $x = 1$ into the assumed linear function to find the value of $m$:
$$ f(1) = m \cdot 1 = m. $$
Thus,
$$ m = 7, $$
which leads us to the function
$$ f(x) = 7x. $$
Now, we need to calculate the sum
$$ \sum_{r=1}^{n} f(r). $$
Substituting $f(r) = 7r$ into the summation:
$$ \sum_{r=1}^{n} f(r) = \sum_{r=1}^{n} 7r = 7 \sum_{r=1}^{n} r. $$
The sum of the first $n$ positive integers is given by the formula
$$ \sum_{r=1}^{n} r = \frac{n(n+1)}{2}. $$
Thus,
$$ \sum_{r=1}^{n} f(r) = 7 \left(\frac{n(n+1)}{2}\right) = \frac{7 n (n+1)}{2}. $$
Therefore, the correct option is (\mathbf{A}):
$$ \frac{7n(n+1)}{2}. $$
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