If f(x) satisfies the relation f(x+y)=f(x)+f(y) for all x, y∈R and f(1) = 5, then

f(2)=f(1+1)=2f(1)=10f(3)=f(2+1)=f(2)+f(1)=10+5=15Then, f(n) = 5nor ∑r=1m f(r)=5∑r=1m r=5m(m+1)2Replacing y by -x, we get f(0)=f(x)+f(−x)Also, putting x = y = 0, we getf(0)=f(0)+f(0) or f(0)=0So, f(x)+f(−x)=0. Hence, the function is odd.