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→∞).

See repost

http://www.jiskha.com/display.cgi?id=1305228142

To determine the long-term behavior of the sequence, let's analyze the expression for un.

un = 2×(-0.5)^n + 3

As n approaches infinity, the term (-0.5)^n will become smaller and smaller, eventually approaching zero. Therefore, the dominant term becomes 3.

So, as n approaches infinity, un will approach 3.

Therefore, the correct option is C. un approaches 3 as n approaches infinity (un → 3 as n → ∞).

To determine the long-term behavior of the sequence, we need to evaluate the limit of the sequence as n approaches infinity.

The given sequence is defined as un = 2×(−0.5)n + 3.

As n approaches infinity, the value of (-0.5)^n becomes smaller and smaller. Since (-0.5) is a negative number, raising it to larger powers makes it closer to zero but with alternating signs.

To determine the behavior of the sequence, we can consider the two terms in the sequence: 2×(−0.5)n and 3.

Since (-0.5)^n approaches 0 as n approaches infinity, the term 2×(−0.5)n approaches 0.

The constant term 3 remains the same regardless of n.

Combining these, the sequence un approaches 3 as n approaches infinity, which means un → 3 as n → ∞.

Therefore, the correct option that best describes the long-term behavior of the sequence is:

C) un approaches 3 as n → ∞ (that is, un → 3 as n → ∞).