Question

If f : R ${\displaystyle \to }$ R satisfies f(x + y) = f(x) + f(y), for all x, y ∈ R and f(1) = 7, then ${\displaystyle \sum _{r=1}^{n}f\left(r\right)}$ is ______.

Options

• ${\displaystyle \frac{7n(n+1)}{2}}$

• ${\displaystyle \frac{7n}{2}}$

• ${\displaystyle \frac{7(n+1)}{2}}$

• 7n + (n + 1)

Answer 1

If f: R ${\displaystyle \to }$ R satisfies f(x + y) = f(x) + f(y), for all x, y ∈ R and f(1) = 7, then ${\displaystyle \sum _{r=1}^{n}f\left(r\right)}$ is ${\displaystyle \underline{\frac{7n(n+1)}{2}}}$.

Explanation:

Let x = 0 = y 

Therefore, f(0) = 0 and x = 1, y = 0

Therefore, f(1 + 0) = f(1) + f(0) = 7  ...(Given)

x = 1, y = 1 

Therefore, f(1 + 1) = 2f(1) = 2(7)

Therefore, f(2) = 2(7)

x = 1, y = 2

Therefore, f(3) = f(1) + f(2)

= 7 + 2(7)

= 3(7)

and so on

∴ ${\displaystyle \sum _{r=1}^{n}f\left(r\right)}$ = f(1) + f(2) + f(3) + .... + f(n)

= 7(1 + 2 + 3 + ... + n)

= ${\displaystyle \frac{7n(n+1)}{2}}$