The sequence $\left\{ a_n \right\}$ satisfies $a_1 = 4$ and $$a_{n+1} = n \cdot 2^n + \sum_{k=1}^{n} \frac{a_k}{k} \quad (n \geq 1)$$ The following is the process of finding the general term $a_n$. From the given equation, $$a_n = (n-1) \cdot 2^{n-1} + \sum_{k=1}^{n-1} \frac{a_k}{k} \quad (n \geq 2)$$ Therefore, for natural numbers $n \geq 2$, $$a_{n+1} - a_n = \text{(가)} + \frac{a_n}{n}$$ so $$a_{n+1} = \frac{(n+1)a_n}{n} + \text{(가)}$$ If $b_n = \frac{a_n}{n}$, then $$b_{n+1} = b_n + \frac{(\text{가})}{n+1} \quad (n \geq 2)$$ and since $b_2 = 3$, $$b_n = \text{(나)} \quad (n \geq 2)$$ Therefore, $$a_n = \begin{cases} 4 & (n = 1) \\ n \times (\text{나}) & (n \geq 2) \end{cases}$$ If the expressions for (가) and (나) are $f(n)$ and $g(n)$, respectively, what is the value of $f(4) + g(7)$? [4 points] (1) 90 (2) 95 (3) 100 (4) 105 (5) 110
The sequence $\left\{ a_n \right\}$ satisfies $a_1 = 4$ and
$$a_{n+1} = n \cdot 2^n + \sum_{k=1}^{n} \frac{a_k}{k} \quad (n \geq 1)$$
The following is the process of finding the general term $a_n$.
From the given equation,
$$a_n = (n-1) \cdot 2^{n-1} + \sum_{k=1}^{n-1} \frac{a_k}{k} \quad (n \geq 2)$$
Therefore, for natural numbers $n \geq 2$,
$$a_{n+1} - a_n = \text{(가)} + \frac{a_n}{n}$$
so
$$a_{n+1} = \frac{(n+1)a_n}{n} + \text{(가)}$$
If $b_n = \frac{a_n}{n}$, then
$$b_{n+1} = b_n + \frac{(\text{가})}{n+1} \quad (n \geq 2)$$
and since $b_2 = 3$,
$$b_n = \text{(나)} \quad (n \geq 2)$$
Therefore,
$$a_n = \begin{cases} 4 & (n = 1) \\ n \times (\text{나}) & (n \geq 2) \end{cases}$$
If the expressions for (가) and (나) are $f(n)$ and $g(n)$, respectively, what is the value of $f(4) + g(7)$? [4 points]\\
(1) 90\\
(2) 95\\
(3) 100\\
(4) 105\\
(5) 110