For $n\in\mathbb{N}$, let $\varphi:\mathbb{R}\to\mathbb{R}$ be a $4n-$ periodic function s.t. $$\varphi(t)=n-|n-t|,\quad -n\leq t\leq 3n$$ For each $j\in\mathbb{Z}$ define $$c_j=\int_0^1\varphi(4n t)e^{-2\pi ij t}dt$$
(i) Show that $$\sum_{j\in\mathbb{Z}}|c_j|=n.$$
(ii) Let $Q_n(x)=\sum_{k=-n}^na_ke^{2\pi ik x}$ be a trigonometric polynomial of degree $n$, show that $$Q_n'(x)=2\pi i\sum_{j\in\mathbb{Z}}c_jQ_n\left(x+\frac{j}{4n}\right)$$
For (i), I got $$c_j=\begin{cases}\frac{4ni}{\pi^2j^2}\sin\left(\frac{\pi j}{2}\right), &j\text{ odd}\\ 0, &j\text{ even}\end{cases}$$ thus (i) is quite straight forward.
I've been really struggling on (ii) for a while. I wrote $$\begin{aligned} c_jQ_n\left(x+\frac{j}{4n}\right)&=c_j\sum_{k=-n}^{n}a_ke^{2\pi ikx}e^{\frac{\pi ikj}{2n}}\\ &=\frac{2n}{\pi^2 j^2}(e^{\frac{\pi ij}{2}}-e^{-\frac{\pi ij}{2}})\sum_{k=-n}^{n}a_ke^{2\pi ikx}e^{\frac{\pi ikj}{2n}}\\ &=\frac{2n}{\pi^2 j^2}\sum_{k=-n}^{n}a_ke^{2\pi ikx}(e^{\frac{\pi i(k+n)j}{2n}}-e^{\frac{\pi i(k-n)j}{2n}}) \end{aligned}$$ for all odd $j$, so $$\sum_{j\in\mathbb{Z}}c_jQ_n\left(x+\frac{j}{4n}\right)=\sum_{k=-n}^{n}a_ke^{2\pi ikx}\sum_{j\in\mathbb{Z}, j\text{odd}}\frac{2n}{\pi^2 j^2}(e^{\frac{\pi i(k+n)j}{2n}}-e^{\frac{\pi i(k-n)j}{2n}})$$ At this point I am stuck. Is there any trick to prove the result? Thanks in advance.