Form or asymptotic behaviour of $T(n) =2T(n-1)+n$

Question

$T(n) =$ if $n=1$, then time execution is $1$, if $n \geq 2$ then $2T(n-1)+n$

The options are:

1. $T(n) = 2^{n+1} - n - 2$
2. $T(n) = O(n2^n)$
3. $T(n) = \Omega(n)$
4. $T(n) = \theta(2^n)$

Thanks.

Answer 1

Ignoring the choices you have for the time being, here is how we could start out.

Define $S(n)=2^{-n}T(n)$, then, after multiplying by $2^{-n}$, the recurrence becomes $$ S(n)=S(n-1)+\frac{n}{2^n}\tag{1} $$ Equation $(1)$ tells us that $S(n)$ is the finite sum $$ c+\sum_{k=1}^n\frac{k}{2^k}\tag{2} $$ where $c$ is a constant chosen so that $S(1)=\frac12$.