Answer
No
Work Step by Step
We want to check whether the closed‐form formula
\[
a_n \;=\; (n-1)^2
\]
actually holds for all integers \(n \ge 1\). Let’s label them:
\[
a_k \;=\; 2\,a_{k-1} \;+\; (k - 1),
\quad a_1 = 0,
\quad k \ge 2.
\]
We test whether \(a_n = (n-1)^2\) works.
The recursion is
\[
a_k = 2\,a_{k-1} + (k - 1),
\quad
a_1 = 0.
\]
Check the first few terms and compare to \((n-1)^2\):
\(n = 1\):
- Recursion says \(a_1 = 0\).
- Proposed formula: \((1-1)^2 = 0.\)
These match for \(n=1\).
\(n = 2\):
- From recursion:
\[
a_2 = 2\,a_1 + (2 - 1) = 2 \cdot 0 + 1 = 1.
\]
- Formula: \((2-1)^2 = 1.\)
Matches for \(n=2\).
\(n = 3\):
- Recursion:
\[
a_3 = 2\,a_2 + (3 - 1) = 2 \cdot 1 + 2 = 4.
\]
- Formula: \((3-1)^2 = 4.\)
Matches for \(n=3\).
**\(n = 4\)**:
- Recursion:
\[
a_4 = 2\,a_3 + (4 - 1)
= 2 \cdot 4 + 3
= 8 + 3
= 11.
\]
- Formula: \((4-1)^2 = 9.\)
They do **not** match at \(n=4\). Indeed, the recursion yields \(a_4 = 11\), while \((4-1)^2 = 9\). Hence, after \(n=3\), the two expressions diverge. Therefore,
\[
\boxed{\text{The sequence does not satisfy }a_n = (n-1)^2.}
\]
