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spectrojamz

Answer:

Answer is Option B.

That is the nth term of the sequence.

1/8 + 1/16 + 1/32....

You notice that all the numerators are the same. Which is (1).

so we have to both about getting the nth term of the denominator

Since all the denominator are divisible by 2... and Your intuition should tell you that they'd multiples of 2

So Lets go with 2^n

2^n

let n be all Integers

let take 1,2,3

2¹ = 2

2² = 4

2³ = 8

Now Observe that

If you multiply all these by 4... We will have all the values of the denominator

which are 4,16,32

So thats our nth term

1/4(2^n)

as n approaches positive Infinity.

The expression is [tex]\lim_{n \to \infty} \frac{1}{4(2^{n}) }[/tex] , so option (B) is the correct expression.

What is series expansion?

"A series expansion is an expansion of a function into a series or infinite sum. It is a method for calculating a function that cannot be expressed by just elementary operators (addition, subtraction, multiplication, and division)".

For the given situation,

The series is given as [tex]\frac{1}{8} +\frac{1}{16}+ \frac{1}{32}+ \frac{1}{64}+ \frac{1}{128}+......[/tex]

In the series, all the numerators are 1 and the denominators are even numbers so multiples of 2.

Now, we examine the denominators alone.

[tex]8,16,32,64,128[/tex]

⇒ Multiples of 2. So we get [tex]2^{n}[/tex] multiplied by some integers.

Put [tex]n=1,[/tex]

⇒ [tex]2^{1}[/tex] × [tex]4=8[/tex]

Put [tex]n=2,[/tex]

⇒ [tex]2^{2}[/tex] × [tex]4=8[/tex]

Put [tex]n=3,[/tex]

⇒ [tex]2^{3}[/tex] × [tex]4=32[/tex]

Put [tex]n=4,\\[/tex]

⇒ [tex]2^{4}[/tex] × [tex]4=64[/tex]

Put [tex]n=5[/tex],

⇒ [tex]2^{5}[/tex] × [tex]4=128[/tex]

Thus [tex]2^{n}[/tex] expression becomes [tex]2^{n}(4)[/tex] .

Hence we can conclude that the expression is [tex]\lim_{n \to \infty} \frac{1}{4(2^{n}) }[/tex] . So option (B) is the correct expression.

Learn more about the series expansion here

https://brainly.com/question/14874506

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