Play with functions 3

$f(0+0+1)=(\sqrt{f(0)}+\sqrt{f(0)})^2$

so $f(1)=2^2$ . Similarly $f(n)=(n+1)^2$ (can be proved by induction)

Putting $x=y$ in $f(x+y+1) = \left(\sqrt{f(x)} + \sqrt{f(y)}\right)^2$ , we have:

$\begin{aligned} f(2x+1) & = 4f(x) & \small \color{#3D99F6} \text{Putting }x=0 \\ f(1) & = 4f(0) = 4 & \small \color{#3D99F6} \text{Given that }f(0) = 1 \\ f(3) & = 4f(1) = 4^2 & \small \color{#3D99F6} \text{for }x = 1 \\ f(7) & = 4f(3) = 4^3 & \small \color{#3D99F6} \text{for }x = 3 \\ f(15) & = 4f(7) = 4^4 & \small \color{#3D99F6} \text{for }x = 7 \end{aligned}$

$\begin{aligned} \implies f(2^u-1) & = 4^u & \small \color{#3D99F6} \text{where }u \text{ is a non-negative integer.} \\ f(x) & = 4^{\log_2(x+1)} & \small \color{#3D99F6} \text{Let }x = 2^u - 1 \implies u = \log_2(x+1) \\ & = 2^{2\log_2(x+1)} \\ \implies f(x) & = (x+1)^2 \end{aligned}$

Let us prove that $f(x)=(x+1)^2$ is true for all real $x$ .

$\begin{aligned} f(x+y+1) & = \left(\sqrt{f(x)} + \sqrt{f(y)}\right)^2 & \small \color{#3D99F6} \text{Replace }x \text{ and }y \text{ with }\frac {x-1}2 \\ \implies f(x) & = 4f\left(\frac {x-1}2\right) \\ & = 4\left(\frac {x-1}2-1\right)^2 \\ & = (x+1)^2 \blacksquare \end{aligned}$

Therefore, $f(2014) = 2015^2 = 4060225$ and its sum of digits is $\boxed{19}$ .

by inducation,we get f(n)=sqr(n+1).then f(2014)=sqr(2014+1)=4060225 ans:4+0+6+0+2+2+5=19