1. Given functional equation

We have $f(x)+f(x+k)=n \quad \text{for all } x\in \mathbb R,$ where $k>0$ and $n$ is a positive integer.

We need to compute $I_1=\int_0^{4nk} f(x)\,dx, \qquad I_2=\int_{-k}^{3k} f(x)\,dx.$

2. A useful consequence of the functional equation

Replace $x$ by $x+k$: $f(x+k)+f(x+2k)=n.$

From the original equation, $f(x+k)=n-f(x).$ So, $f(x+2k)=n-f(x+k)=n-(n-f(x))=f(x).$

Hence, $\boxed{f(x+2k)=f(x)}$ for all $x$. Thus $f$ is periodic with period $2k$.

3. Integral over one period

Let $J=\int_0^{2k} f(x)\,dx.$

Using the given relation, $\int_0^k \big(f(x)+f(x+k)\big)\,dx=\int_0^k n\,dx=nk.$

But $\int_0^k f(x+k)\,dx=\int_k^{2k} f(t)\,dt.$ So, $\int_0^k f(x)\,dx+\int_k^{2k} f(t)\,dt=nk.$ Therefore,

So, $\boxed{\int_0^{2k} f(x)\,dx=nk.}$

4. Evaluate $I_1$

Since $f$ has period $2k$, the interval $[0,4nk]$ contains $\frac{4nk}{2k}=2n$ full periods.

Therefore, $I_1=2n\int_0^{2k} f(x)\,dx=2n(nk)=2n^2k.$ So, $\boxed{I_1=2n^2k.}$

5. Evaluate $I_2$

We have $I_2=\int_{-k}^{3k} f(x)\,dx.$ The interval length is $3k-(-k)=4k=2(2k),$ which is exactly two periods.

Hence, $I_2=2\int_0^{2k} f(x)\,dx=2(nk)=2nk.$ So, $\boxed{I_2=2nk.}$

(You can also verify by splitting as $[-k,k]$ and $[k,3k]$, each of length $2k$.)

6. Check the options

Now, $I_1+nI_2=2n^2k+n(2nk)=2n^2k+2n^2k=4n^2k.$ Thus, $\boxed{I_1+nI_2=4n^2k.}$

So option C is correct.

Let us quickly reject others:

• A: $I_1+2I_2=2n^2k+4nk,$ not always equal to $4nk$.

• B: $I_1+2I_2=2n^2k+4nk,$ not always equal to $2nk$.

• D: $I_1+nI_2=4n^2k,$ not $6n^2k$.

7. Conclusion

The correct option is $\boxed{\text{C}}.$