Mathematics & Numerical Analysis

Vieta's formulas for quadratic polynomials:
For a quadratic polynomial $ax^2 + bx + c$:
Sum of roots $\alpha + \beta = -\frac{b}{a}$.
Product of roots $\alpha \beta = \frac{c}{a}$.
Here, $a = 1$, $b = -2$, and $c = -8$:
Sum $= -\frac{-2}{1} = 2$.
Product $= \frac{-8}{1} = -8$.

Option breakdown:

• $2, 8$ has the wrong sign for the product.
• $-2, 8$ has reversed signs for both sum and product.
• $-2, -8$ has the wrong sign for the sum.
• $2, -8$ is the exact sum and product of the roots.

Convergence analysis:

• For $(a_n) = \frac{n+1}{n} = 1 + \frac{1}{n}$:
  $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = 1 + 0 = 1$$
  Since the limit exists and is finite, $(a_n)$ converges to 1.
• For $(b_n) = (-1)^n$:
  The terms alternate: $-1, 1, -1, 1, \dots$. As $n \to \infty$, $(b_n)$ has two subsequential limit points (-1 and +1), so $\lim_{n\to\infty} (-1)^n$ does not exist. Hence $(b_n)$ is oscillatory (not convergent).

Convergence of the sequence $(a_n) = \frac{1}{n^r}$:
If $r \gt 0$: $\lim_{n \to \infty} \frac{1}{n^r} = 0$, so $(a_n)$ converges to 0.
If $r = 0$: $a_n = \frac{1}{n^0} = 1$ for all $n$, so $(a_n)$ is a constant sequence converging to 1.
If $r \lt 0$: Let $r = -k$ where $k \gt 0$. Then $a_n = n^k \to \infty$ as $n \to \infty$, which diverges.