If f:R→R satisfies f(x+y)=f(x)+f(y), for all x,y∈R, f(1)=32, then ∑f(r)r=1nis
32n(n+1)
3n+12
3nn+14
3n4
Finding ∑f(r)r=1n:
We have: f1=32 , f(x+y)=f(x)+f(y)
∵f(1)=32
∴f(2)=f(1+1)=f(1)+f(1)=32+32=62
Similarly,
∴f(3)=f(2+1)=f(2)+f(1)=62+32=92
∴f(4)=f(1+3)=f(1)+f(3)=32+92=122 and so on.
∴f(n)=32+(n-1)32
Now, ∑f(r)r=1n=f(1)+f(2)+f(3)+....f(n)
=32+62+92+122+....+(n-1)32
=32(1+2+3+4+.....n)=3n(n+1)4
Hence, the correct option is an option (c).