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Step-by-Step Solution
Step 1: Understand the Given Functional Equation
We are given a continuous function $f : \mathbb{R} \to \mathbb{R}$ satisfying
$f(x) + f(x + k) = n \text{ for all } x \in \mathbb{R},$
where $k > 0$ and $n$ is a positive integer.
Step 2: Find the Period of the Function
By substituting $x \mapsto x + k$ in the given relationship, we get:
\[
f(x + k) + f(x + 2k) = n.
\]
We already have $f(x) + f(x + k) = n,$ so by comparing both equations, we deduce:
\[
f(x + 2k) = f(x).
\]
Hence, $f(x)$ has period $2k.$
Step 3: Express $I_1$ in Terms of One Period
The integral
\[
I_1 = \int_{0}^{4nk} f(x)\,dx
\]
can be seen as integrating over $2n$ periods, since one period is $2k$. Thus,
\[
I_1
= \int_{0}^{4nk} f(x)\,dx
= 2n \int_{0}^{2k} f(x)\,dx.
\]
Next, split the integral $\int_{0}^{2k} f(x)\,dx$ into two integrals from $0$ to $k$ and from $k$ to $2k$:
\[
\int_{0}^{2k} f(x)\,dx
= \int_{0}^{k} f(x)\,dx + \int_{k}^{2k} f(x)\,dx.
\]
Step 4: Use the Functional Equation
For the second integral, make the substitution $x = t + k$, $dx = dt$. Then
\[
\int_{k}^{2k} f(x)\,dx
= \int_{0}^{k} f(t + k)\,dt.
\]
But we know from $f(x) + f(x + k) = n$ that $f(t + k) = n - f(t).$
So
\[
\int_{0}^{k} f(t + k)\,dt
= \int_{0}^{k} \bigl[n - f(t)\bigr]\,dt
= n\int_{0}^{k} 1\,dt - \int_{0}^{k} f(t)\,dt
= nk - \int_{0}^{k} f(t)\,dt.
\]
Hence,
\[
\int_{0}^{2k} f(x)\,dx
= \int_{0}^{k} f(x)\,dx + \left[nk - \int_{0}^{k} f(x)\,dx\right]
= nk.
\]
Therefore,
\[
I_1 = 2n \int_{0}^{2k} f(x)\,dx = 2n \cdot (nk) = 2n^2 k.
\]
Step 5: Compute $I_2$
Similarly,
\[
I_2 = \int_{-k}^{3k} f(x)\,dx.
\]
Observe that the interval $[-k,3k]$ has length $4k,$ which is two periods (since the period is $2k$). We can shift the limits to make use of the periodicity. Noting that shifting a definite integral by a period does not change its value, we can write:
\[
I_2 = \int_{0}^{4k} f(x)\,dx.
\]
But a single period $[0,2k]$ integral is $\int_{0}^{2k} f(x)\,dx = nk$, so over $[0,4k]$, it is twice that:
\[
I_2 = 2 \cdot \left(\int_{0}^{2k} f(x)\,dx\right) = 2(nk) = 2nk.
\]
Step 6: Evaluate $I_1 + n I_2$
Finally,
\[
I_1 + n I_2 = 2n^2 k + n(2nk) = 2n^2 k + 2n^2 k = 4n^2 k.
\]
Thus, the correct relation is
\[
I_1 + n I_2 = 4n^2 k.
\]
Answer
The correct answer is:
\[
I_1 + n I_2 = 4n^2 k.
\]
