Sequence and series (4)

The first few $n$ indicate that $x_n = an^2 + bn$ . Let us prove by induction that the claim is true for all $n\ge 0$ .

For $n=0$ , it is given that $x_0 = 0 = a(0^2)+b(0)$ . So the claim is true for $n=0$ . Assuming the claim is true for $n$ , then

$\begin{aligned} x_{n+1} & = x_n + a + \sqrt{b^2+4ax_n} \\ & = an^2 + bn + a + \sqrt{b^2 + 4a(an^2+bn)} \\ & = a(n^2+1) + bn + \sqrt{b^2+4abn+4a^2n^2} \\ & = a(n^2+1) + bn + b+2an \\ & = a(n^2 + 2n+1) + b(n+1) \\ & = a(n+1)^2 + b(n+1) \end{aligned}$

The claim is also true for $n+1$ , therefore it is true for all $n \ge 0$ . And $\displaystyle \lim_{n \to \infty} \frac {x_n}{n^2} = \lim_{n \to \infty} \frac {an^2+bn}{n^2} = \lim_{n \to \infty} \left(a + \frac bn\right) = \boxed a$ .