If f: R → R satisfies f (x + y) = f (x) + f (y), for all x, y ∈ R and f (1) = 7, then displaystyle ∑r=1n f(r) is

Question

If f: R → R satisfies f (x + y) = f (x) + f (y), for all x, y ∈ R and f (1) = 7, then displaystyle ∑r=1n f(r) is (A)  (7n/2) (B)  (7(n+1)/2) (C)  7n(n+1) (D) 7 (n(n+1)/2)

Tardigrade Admin · Accepted Answer

f (1) = 7 f (1 + 1) = f (1) + f (1) f (2) = 2 Ã 7 only f (3) = 3 Ã 7 displaystyle ∑r=1n f(r) = 7 (1 + 2 + â¦â¦â¦ + n) =7 (n(n+1)/2).

Q. If $f : R \to R$ satisfies $f (x + y) = f (x) + f (y)$ , for all $x, y \in R$ and $f (1) = 7$ , then $r = 1 \sum n f (r)$ is

$\frac{7 n}{2}$

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

$7 n (n + 1)$

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

$f (1) = 7$
$f (1 + 1) = f (1) + f (1)$
$f (2) = 2 \times 7$
only $f (3) = 3 \times 7$
$r = 1 \sum n f (r) = 7 (1 + 2 + \dots\dots\dots + n)$
$= 7 \frac{n ( n + 1 )}{2} .$

Q. If f:R→R satisfies f(x+y)=f(x)+f(y), for all x,y∈R and f(1)=7, then r=1∑n​f(r) is

27n​

27(n+1)​

7n(n+1)

72n(n+1)​