Question

If f:R-R is defined as f(x + y) = f(x) + f(y) ¥x, y € R and f (1) = 7,
then find Ef(r).
r=1​

Answer 1

To find :

$\displaystyle \sf \sum \limits_{ r=1}^{7} f(r) = ?$

Given that,

$\sf \: f(x + y) = f(x) + f(y) \: and \: f(1) = 7$

Now,

$\displaystyle \sf \sum \limits_{ r=1}^{7} f(r) = f(1) + f(2) + \cdots \: + f(7)$

Putting x = y = 1,

$\sf \: f(2) = f(1 + 1) \\ \\ \longrightarrow \sf \: f(2) = f(1) + f(1) \\ \\ \longrightarrow \sf \: f(2) = 14$

Putting x = 2 and y = 1,

$\sf \: f(3) = f(2 + 1) \\ \\ \longrightarrow \sf \: f(3) = f(2) + f(1) \\ \\ \longrightarrow \sf \: f(3) =21$

Putting x = y = 2,

$\sf \: f(4) = f(2 + 2) \\ \\ \longrightarrow \sf \: f(4) = f(2) + f(2) \\ \\ \longrightarrow \sf \: f(4) =28$

Putting x = 4 and y = 1,

$\sf \: f(5) = f(4 + 1) \\ \\ \longrightarrow \sf \: f(5) = f(4) + f(1) \\ \\ \longrightarrow \sf \: f(5) = 35$

Putting x = 5 and y = 1,

$\sf \: f(6) = f(5 + 1) \\ \\ \longrightarrow \sf \: f(6) = f(5) + f(1) \\ \\ \longrightarrow \sf \: f(6) = 42$

Putting x = 4 and y = 3,

$\sf \: f(7) = f(4 + 3) \\ \\ \longrightarrow \sf \: f(7) = f(4) + f(3) \\ \\ \longrightarrow \sf \: f(7) = 49$

Now,

$\implies \displaystyle \sf \sum \limits_{ r=1}^{7} f(r) = 7 + 14 + 21 + 28 + 35 + 42 + 49 \\ \\ \implies \displaystyle \boxed{ \boxed{ \sf \sum \limits_{ r=1}^{7} f(r) = 196}}$

Answer 2

Correct question= If f : R → R satisfies f (x + y) = f (x) + f (y), for all x,

Answer:-⬇️

If f : R → R satisfies f (x + y) = f (x) + f (y), for all x,