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[br]\begin{array}{rcl}[br]\displaystyle\sum_{k=0}^{n} 2^k(n+1-k) &=& (n+1)\displaystyle\sum_{k=0}^{n} 2^k - 2\displaystyle\sum_{k=0}^{n} k2^{k-1} \\[br]&=& (n+1)(2^{n+1}-1) - 2(1-(n+1)2^n+n2^{n+1})\\[br]\\[br]&=& 2^{n+2}-n-3[br]\end{array}
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\displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{ t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1
\displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi} {(2k)!}-\frac{x}{(2k+1)!}\right)