Here, $T(1)=1$ and
$\begin{align} T(n)&=2T(n-1)+n\\&=2^2T(n-2)+2(n-1)+n;~[\text{Putting }T(n-1)=2T(n-2)+(n-1)]\\&=2^3T(n-3)+2^2(n-2)+2(n-1)+n;~[\text{Doing so}]\\&=\cdots\\&=2^{n-1}T(n-(n-1))+2^{n-2}\{n-(n-2)\}+2^{n-3}\{n-(n-3)\}+\cdots+2(n-1)+n\\&=2^{n-1}T(1)+2^{n-2}(2)+2^{n-3}(3)+\cdots+2^{1}(n-1)+2^{0}n\end{align}$
Now putting $T(1)=1$ as given,
$\begin{align}\therefore T(n)&=2^{n-1}(1)+2^{n-2}(2)+2^{n-3}(3)+\cdots+2^{1}(n-1)+2^{0}n \tag{i} \\ \Rightarrow 2T(n)&=2^n(1)+2^{n-1}(2)+2^{n-2}(3)+\cdots+2^2(n-1)+2^{1}n \tag{ii}\end{align} $
$\mathrm{no(ii)}-\mathrm{no(i)}\Rightarrow\\ \begin{align}T(n)&=2^n+2^{n-1}(2-1)+2^{n-2}(3-2)+\cdots+2^1(n-(n-1))-2^{0}n\\&=(2^{n}+2^{n-1}+2^{n-2}+\cdots+2)-n\\&=(2+2^2+2^3+\cdots+2^{n-1}+2^n)-n; ~[\text{Rearranging}]\\&=\frac{2(2^n-1)}{2-1}-n;~[\scriptsize\because a+ar+ar^2+\cdots+ar^n=\frac{a(r^{n+1}-1)}{r-1}\text{ as Geometric Series}]\\&=2^{n+1}-2-n \end{align}$
$\therefore T(n)=2^{n+1}-n-2$.
So the correct answer is A.