Answer
\[
\boxed{f_k = 2^{k+1} - 3.}
\]
Work Step by Step
We have the recurrence
\[
f_k = f_{k-1} + 2^k \quad \text{for } k \ge 2,
\quad \text{with} \quad f_1 = 1.
\]
To guess the closed form, let us iterate (unroll) the recurrence:
\[
f_k
= f_{k-1} + 2^k
= \bigl(f_{k-2} + 2^{k-1}\bigr) + 2^k
= \dots
= f_1 + 2^2 + 2^3 + \cdots + 2^k.
\]
Since \(f_1 = 1\), we have
\[
f_k = 1 + \sum_{j=2}^k 2^j.
\]
We can sum the geometric series:
\[
\sum_{j=2}^k 2^j = 2^2 + 2^3 + \cdots + 2^k.
\]
Recall the standard geometric sum formula:
\[
\sum_{j=0}^k 2^j = 2^{k+1} - 1.
\]
Thus,
\[
\sum_{j=2}^k 2^j
= \Bigl(\sum_{j=0}^k 2^j\Bigr) - \bigl(2^0 + 2^1\bigr)
= \bigl(2^{k+1} - 1\bigr) - (1 + 2)
= 2^{k+1} - 1 - 3
= 2^{k+1} - 4.
\]
Therefore,
\[
f_k
= 1 + \bigl(2^{k+1} - 4\bigr)
= 2^{k+1} - 3.
\]
