Solve the recurrence $T(n) = 2T(n-1) + n$ Solve the recurrence  $T(n) = 2T(n-1) + n$
where $T(1) = 1$ and $n\ge 2$.
The final answer is $2^{n+1}-n-2$
Can anyone arrive at the solution?

\begin{align}
T(n) & = 2 T(n-1) + n = 2(2T(n-2) + n-1) + n = 4T(n-2) + 2(n-1) + n\\
& = 8 T(n-3) + 4(n-2) + 2(n-1) + n = 2^k T(n-k) + \sum_{j=0}^{k-1} 2^j (n-j)\\
& = 2^{n-1} T(1) + \sum_{j=0}^{n-2}2^j (n-j) = 2^{n-1} + \sum_{j=0}^{n-2}2^j (n-j)
\end{align}
\begin{align}
\sum_{j=0}^{n-2}2^j (n-j) & =  n \sum_{j=0}^{n-2}2^j - \sum_{j=0}^{n-2} j2^j = n(2^{n-1}-1) - \dfrac{n \cdot 2^n - 3 \cdot 2^n + 4}2\\
& = n(2^{n-1}-1) - (n \cdot 2^{n-1} -3 \cdot 2^{n-1} + 2) = 3 \cdot 2^{n-1} -n - 2
\end{align}
Hence,
$$T(n) = 2^{n-1} + 3 \cdot 2^{n-1} -n - 2 = 2^{n+1} - n - 2$$
EDIT (Adding details)
First note that $\displaystyle \sum_{j=0}^{n-2}2^j$ is sum of a geometric progression and can be summed as shown below.$$\sum_{j=0}^{k} x^j = \dfrac{x^{k+1} -1}{x-1}$$
$\displaystyle \sum_{j=0}^{n-2} j2^j$ is a sum of the form $\displaystyle \sum_{j=0}^{k} jx^j$
$$\sum_{j=0}^{k} jx^j = x \sum_{j=0}^{k} jx^{j-1} = x \dfrac{d}{dx} \left( \sum_{j=0}^k x^j\right) = x \dfrac{d}{dx} \left( \dfrac{x^{k+1} - 1}{x-1}\right) = x \left( \dfrac{kx^{k+1} - (k+1) x^k +1}{(x-1)^2} \right)$$
