Let f: R → R be continuous function satisfying f(x) + f(x + k) = n...

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,

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

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

Now, ${I}_{1} = {∫}_{0}^{4nk}f\left(x\right)dx = 2n{∫}_{0}^{2k}f\left(x\right)dx$

${I}_{2} = {∫}_{- k}^{3k}f\left(x\right)dx = 2{∫}_{0}^{2k}f\left(x\right)dx$

Now integrating both side of $f\left(x\right) + f\left(x + k\right) = n$ with limit ${∫}_{0}^{k}$ we get,

$⇒ {∫}_{0}^{k}f\left(x\right)dx + {∫}_{0}^{k}f\left(x + k\right)dx = nk$

$⇒ {∫}_{0}^{k}f\left(x\right)dx + {∫}_{k}^{2k}f\left(x\right)dx = nk$

$⇒ {∫}_{0}^{2k}f\left(x\right)dx = nk$

$⇒ {I}_{1} = 2{n}^{2}k, {I}_{2} = 2nk$

$⇒ {I}_{1} + n{I}_{2} = 4{n}^{2}k$