Finding the sum of n terms $S_n$ starting from sigma $k=0$ $$\sum_{k=0}^{n} ((4k-3)\cdot 2^k)+4=(2^{n+3}+4)n-7\cdot2^{n+1}+15$$
How? I've tried everything but i don't see it. Any equivalent solutions are also welcome, thanks. 

You can exploit a telescoping sum:
$$
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)=(n+1)2^{n+1},
$$
because you also have:
$$
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)=
\sum_{k=0}^n\bigl(k2^k-2^{k+1}\bigr)
$$
so that
$$
\sum_{k=0}^n  k2^k=
\sum_{k=0}^n\bigl((k+1)2^{k+1}-k2^k\bigr)+\sum_{k=0}^n2^{k+1}
=(n+1)2^{n+1}+2(2^{n+1}-1)
$$
