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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