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 \(n \ge 1\).
\[
a_k \;=\; \bigl(a_{k-1} + 1\bigr)^2,
\quad a_1 = 0,
\quad k \ge 2.
\]
We test whether \(a_n = (n-1)^2\) works.
The recursion is
\[
a_k = (a_{k-1} + 1)^2,
\quad
a_1 = 0.
\]
Again, compare to \((n-1)^2\):
\(n = 1\):
- \(a_1 = 0\).
- Formula: \((1-1)^2 = 0.\)
Matches at \(n=1\).
\(n = 2\):
- Recursion: \(a_2 = (a_1 + 1)^2 = (0+1)^2 = 1.\)
- Formula: \((2-1)^2 = 1.\)
Matches at \(n=2\).
\(n = 3\):
- Recursion: \(a_3 = (a_2 + 1)^2 = (1+1)^2 = 4.\)
- Formula: \((3-1)^2 = 4.\)
Matches at \(n=3\).
**\(n = 4\)**:
- Recursion:
\[
a_4 = (a_3 + 1)^2 = (4 + 1)^2 = 5^2 = 25.
\]
- Formula: \((4-1)^2 = 9.\)
They diverge at \(n=4\). The recursion yields \(a_4 = 25\), while the proposed formula says 9. Therefore,
\[
\boxed{\text{The sequence also does not satisfy }a_n = (n-1)^2.}
\]
