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(C) Given $f(x) + f(x + k) = n$. Replacing $x$ with $x + k$,we get $f(x + k) + f(x + 2k) = n$. Subtracting the two equations,$f(x + 2k) - f(x) = 0$,so

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$I_{1} + nI_{2} = 4n^{2}k$

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(C) Given $f(x) + f(x + k) = n$. Replacing $x$ with $x + k$,we get $f(x + k) + f(x + 2k) = n$. Subtracting the two equations,$f(x + 2k) - f(x) = 0$,so $f(x + 2k) = f(x)$. Thus,$f(x)$ is a periodic function with period $T = 2k$. For a periodic function with period $T$,$\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$. $I_{1} = \int_{0}^{4nk} f(x) dx = \int_{0}^{(2n)2k} f(x) dx = 2n \int_{0}^{2k} f(x) dx$. $I_{2} = \int_{-k}^{3k} f(x) dx = \int_{0}^{4k} f(x) dx = 2 \int_{0}^{2k} f(x) dx$. Now,$\int_{0}^{2k} f(x) dx = \int_{0}^{k} f(x) dx + \int_{k}^{2k} f(x) dx$. Since $f(x + k) = n - f(x)$,$\int_{k}^{2k} f(x) dx = \int_{0}^{k} f(x + k) dx = \int_{0}^{k} (n - f(x)) dx = nk - \int_{0}^{k} f(x) dx$. Therefore,$\int_{0}^{2k} f(x) dx = \int_{0}^{k} f(x) dx + nk - \int_{0}^{k} f(x) dx = nk$. Substituting this back,$I_{1} = 2n(nk) = 2n^{2}k$ and $I_{2} = 2(nk) = 2nk$. Finally,$I_{1} + nI_{2} = 2n^{2}k + n(2nk) = 2n^{2}k + 2n^{2}k = 4n^{2}k$.

Let $f : R \rightarrow R$ be a continuous function satisfying $f(x) + f(x + k) = n$,for all $x \in R$,where $k > 0$ and $n$ is a positive integer. If $I_{1} = \int_{0}^{4nk} f(x) dx$ and $I_{2} = \int_{-k}^{3k} f(x) dx$,then:

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