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. 

Use this important geometric sum :
Lemma 1 
$$A(n)=\sum_{k=0}^{n} 2^k=2^{n+1}-1$$ 
We'll now find another sum :
Lemma 2
$$B(n)=\sum_{k=0}^{n} k 2^k$$
To find this sum simply break it in sums you already know from Lemma 1 like this :
$$B(n)=0+1 \cdot 2^1+2 \cdot 2^2+\ldots n \cdot 2^n=(2^1+2^2+\ldots +2^n)+(2^2+2^3+\ldots +2^n)+\ldots+(2^n)=2A(n-1)+2^2A(n-2)+\ldots +2^nA(0)=2(2^n-1)+2^2(2^{n-1}-1)+\ldots +(2^{n+1}-2^n)=n2^{n+1}-(2+2^2+\ldots+2^n)=n2^{n+1}-(2^{n+1}-2)=(n-1)2^{n+1}+2$$ 
Now your sum is just $$4B(n)-3A(n)+4(n+1)$$ which you can find by plugging in from Lemma 1 and 2.
