Givenf:R→R is defined as f(x+y)=f(x)+f(y),∀x,yinR and f(1)=5 To Find
n
∑
r=1
f(r)= ? Since, f(x+y)=f(x)+f(y) ...(i) Let x=y=1 f(1+1)=f(1)+f(1)=5+5 f(2)=10 Again in Eq. (i), put x=2,y=1 f(2+1)=f(2)+f(1) f(3)=10+5 f(3)=15 Similarly, we can find f(4)=15+5=20 f(5)=25,...f(n)=5n
n
∑
r=1
f(r)=f(I)+f(2)+...+f(n) =5+10+15+...+5n which forms an AP of n-terms with first term (a)=5 and common difference (d)=5 =