A sequence is defined by
un = 2×(−0.5)n + 3 (n = 1,2,3, . . .).
Choose the option that best describes the long-term behaviour of the
sequence.
Options
A un becomes arbitrarily small (that is, un → 0 as n→∞).
B un becomes arbitrarily large and negative (that is, un →−∞
as n→∞).
C un approaches 3 as n→∞ (that is, un → 3 as n→∞).
D un is unbounded and alternates in sign.
E un becomes arbitrarily large and positive (that is, un→∞
as n→∞).
F un approaches 2 as n→∞ (that is, un → 2 as n→∞
Look at the behaviour of the first term:
2*(-0.5)^n (check, this is not what you posted).
We can make the following observations.
As n increases, the sign alternates.
As n->∞ the term approaches zero.
Based on these observations, can you make a choice from the list of possible answers? In the worst case, you should be able to eliminate quite a few choices.
To determine the long-term behavior of the sequence, we can analyze the expression for un as n approaches infinity.
The given sequence is defined as un = 2×(−0.5)n + 3. Let's simplify this expression:
When n is even, (-0.5)n will be positive, giving us:
un = 2 × (0.5) + 3 = 1 + 3 = 4.
When n is odd, (-0.5)n will be negative, giving us:
un = 2 × (-0.5) + 3 = -1 + 3 = 2.
From this, we can see that the sequence alternates between the values 4 and 2 as n increases. Therefore, the sequence is not converging towards a specific value, and it is not becoming arbitrarily small or arbitrarily large (positive or negative) as n approaches infinity.
The correct option is D: un is unbounded and alternates in sign.