4.079 Find the sum $\Sigma=1+2 \cdot 3+3 \cdot 7+\ldots+n\left(2^{n}-1\right)$.
Solution. $\Sigma=1 \cdot\left(2^{1}-1\right)+2 \cdot\left(2^{2}-1\right)+\ldots+n \cdot\left(2^{n}-1\right)=2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n}-$ $-(1+2+3+\ldots+n)$

Let: $S_{n}=2^{i}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+n \cdot 2^{n} \Rightarrow$

$\Rightarrow S_{n-1}=2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+(n-1) 2^{n} \Rightarrow$

$\Rightarrow 2 S_{n-1}=2^{2}+2 \cdot 2^{3}+3 \cdot 2^{4}+\ldots+(n-1) 2^{n} \Rightarrow$
$\Rightarrow S_{n}-2 S_{n-1}=2^{1}+2^{2}(2-1)+2^{3}(3-2)+\ldots+2^{n}(n-(n-1))=$ $=2^{1}+2^{2}+2^{3}+\ldots+2^{n}=2\left(2^{n}-1\right)$

In the end, we have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ S _ { n } - 2 S _ { n - 1 } = 2 ^ { n + 1 } - 2 , } \\
{ S _ { n } - S _ { n - 1 } = n \cdot 2 ^ { n } }
\end{array} \Rightarrow \left\{\begin{array}{l}
S_{n}-2 S_{n-1}=2^{n+1}-2, \\
2 S_{n}-2 S_{n-1}=n \cdot 2^{n+1}
\end{array} \Rightarrow S_{n}=(n-1) 2^{n+1}+2 \Rightarrow\right.\right. \\
& \Rightarrow \Sigma=S_{n}-(1+2+3+\ldots+n)=(n-1) 2^{n+1}+2-\frac{n(n+1)}{2} .
\end{aligned}
$$

Answer: $(n-1) 2^{n+1}+2-\frac{n(n+1)}{2}$.