The given question is:
Suppose \( x_1 = a \) and \( x_{n+1} = x_n + b \). For what values of \( a \) and \( b \) do we have \( x_1 + x_2 + \cdots + x_n = n^2 \quad \text{for all } n \ge 1? \)
Solution:
The recurrence \( x_{n+1} = x_n + b \) with \( x_1 = a \) defines an arithmetic sequence: \( x_n = a + (n-1)b, \quad n \ge 1. \)
Let \(S_n = x_1 + x_2 + \dots + x_n. \)
Then: \( S_n = \sum_{k=1}^n \left[ a + (k-1)b \right] = \sum_{k=1}^n a + b \sum_{k=1}^n (k-1) = na + b \cdot \frac{(n-1)n}{2} \)
Given that \( S_n = n^2 \). Therefore,
\( \begin{aligned} &na + \frac{b}{2}(n^2 – n) = n^2 \\ &\Rightarrow 2na + b(n^2 – n) = 2n^2 \\ &\Rightarrow bn^2 + (2a – b)n = 2n^2 \\ \end{aligned} \)Equating coefficients, we have:
\( \begin{aligned} bn^2 = 2n^2 \Rightarrow b = 2 \\ (2a – b)n = 0 \Rightarrow a = 1 \end{aligned} \)Therefore, the values are: \(a = 1, b = 2 \).
Please let me know in the comments if you find any errors in this solution.