Answer:
∑ (-1)ⁿ⁺³ 1 / (n^½)
∑ (-1)³ⁿ 1 / (8 + n)
Step-by-step explanation:
If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.
Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).
Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).
Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.
Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Answer:
2y + 2z
Step-by-step explanation:
-x + 2z + x + 2y
= (-1 + 1)x + 2y + 2z
= 2y + 2z
Hope this helped!
Answer:
{5, 6, 7}
Step-by-step explanation:
When we have a given relation, the domain is the set of inputs, and the range as the set of the outputs.
so for a function f(x), and a domain {a. b. c}
The range is:
{f(a), f(b), f(c)}
In this case, we have:
f(x) = x + 6
and the domain is {-1, 0, 1}
Then the range is:
{ f(-1), f(0), f(1) }
{-1 + 6, 0 + 6, 1 + 6}
{5, 6, 7}
The correct option is the third one.
