If f:R→R satisfies f(x+y)=f(x)+f(y), for all x,y∈R and f(1)=7 ,then ∑nr=1f(r) is
7n(n+1)2
7n2
7(n+1)2
7n+(n+1)
f(x+y)=f(x)+f(y)Function should be f(x)=mxF(1)=7; ∴ m=7,f(x)=7x∑nr=1f(r)=7∑n1r=7n(n+1)2