Let $f:\mathit{R}\to \mathit{R}$ be continuous function satisfying $f\left(x\right)+f\left(x+k\right)=n$, for all $x\in \mathit{R}$ where $k>0$ and $n$ is a positive integer. If $I_1={\int }_0^{4nk}f\left(x\right)dx$ and $I_2={\int }_{-k}^{3k}f\left(x\right)dx$, then

Given, 

$f\left(x\right)+f\left(x+k\right)=n$ ....(i)

Replacing $x\to x+k$ in above equation we get,

$f\left(x+k\right)+f\left(x+2k\right)=n$ .....(ii)

Subtraction equation (i) from equation (ii) we get,

$\Rightarrow f\left(x\right)=f\left(x+2k\right)$

$f\left(x\right)$ is periodic with period $2k$

Now, $I_1={\int }_0^{4nk}f\left(x\right)dx=2n{\int }_0^{2k}f\left(x\right)dx$

$I_2={\int }_{-k}^{3k}f\left(x\right)dx=2{\int }_0^{2k}f\left(x\right)dx$

Now integrating both side of $f\left(x\right)+f\left(x+k\right)=n$ with limit ${\int }_0^k$  we get,

$\Rightarrow {\int }_0^{k}f\left(x\right)dx+{\int }_0^{k}f\left(x+k\right)dx=nk$

$\Rightarrow {\int }_0^{k}f\left(x\right)dx+{\int }_k^{2k}f\left(x\right)dx=nk$

$\Rightarrow {\int }_0^{2k}f\left(x\right)dx=nk$

$\Rightarrow I_1=2n^{2}k,I_2=2nk$

$\Rightarrow I_1+n{I}_2=4n^{2}k$