某级数的第 $n$ 项为 $2^{n-2} + 3n$。求其前 $N$ 项的和。

1. Split the sum into a geometric series and an arithmetic series.

$S_N = \sum_{n=1}^N (2^{n-2} + 3n) = \sum_{n=1}^N 2^{n-2} + \sum_{n=1}^N 3n$

2. Find the sum of the geometric series component.

The first part is a geometric series with first term $a = 2^{1-2} = \frac{1}{2}$ and common ratio $r = 2$.
$S_{GP} = \frac{\frac{1}{2}(2^N - 1)}{2 - 1} = \frac{1}{2}(2^N - 1)$

3. Find the sum of the arithmetic series component.

The second part is an arithmetic series with first term $a = 3(1) = 3$ and common difference $d = 3$.
$S_{AP} = \frac{N}{2} \left[ 2(3) + (N-1)3 \right] = \frac{N}{2}(3N + 3) = \frac{3}{2}N(N+1)$

4. Combine the sums to give the final expression.

$S_N = \frac{1}{2}(2^N - 1) + \frac{3}{2}N(N + 1)$