If f(x+y+z)=f(x)+f(y)+f(z) with f(1)=1,f(2)=2 and x,y,z εR,thne evaluate \lim_{n\rightarrow\infty }\frac{\sum_{r=1}^{n}{(4r).f(3r)}}{n^3}
21f(x+y+z)=f(x)+f(y)+f(z)
f(0)=0
f(1)=1
f(2)=2
f(3)=f(0+1+2)=3
f(r)=f(0+1+(r-1))
so
f(r)=r
\lim_{n\rightarrow\infty }4(\frac{1.f(3)+2.f(6)+3f(9)..........nf(3n)}{n^{3}}
\lim_{n\rightarrow\infty }4(\frac{1.(3)+2.(6)+3.(9)..........n(3n)}{n^{3}}
general term of series 1.(3)+2.(6)+3.(9)..........n.(3n) be t(3t)
so we have \sum_{t=1}^{t=n}{t.3t}
\sum_{t=1}^{t=n}{3t^{2}}
\frac{3.n(n+1)(2n+1)}{6}
so we hav
\lim_{n\rightarrow \infty }\frac{4(3.n(n+1)(2n+1))}{6n^{3}}
=4
1putting x,y=0;z=1
f(1)=2f(0) +f(1);
f(0)=0;
putting z=0;
f(x+y)=f(x)+f(y)
this is cauchy functional equation;
hence
f(x)=kx
putting x=1
we find k=1;
hence f(x)=x;
f(3r)=3r;
hence the summation reduces to;
\frac{2n(n+1)(2n+1)}{n^{3}}
2\lim_{n\rightarrow infinity}(1+\frac{1}{n})(2+\frac{1}{n})
now them term 1/n will tend to 0;
hence answer is 2*2=4
