For every integer k from 1 to 10, inclusive the "kth" term of a certain sequence is given by [(−1)^(k+1)]∗(1/2^k).
If T is the sum of the first 10 terms in the sequence, then T is:

a) greater than 2.
b) between 1 and 2.
c) between 1/2 and 1.
d) between 1/4 and 1/2.
e) less than 1/4.

Option D

The sum is between 1/4 and 1/2

Step-by-step explanation:

Given the kth-term of the series to be:

T_k = [(-1)^(k + 1)](1/2^k)

For k = 1

T_1 = (-1)²(1/2) = 1/2

For k = 2

T_2 = (-1)³(1/2²) = -1/4

For k = 3

T_3 = (-1)^4 (1/2³) = 1/8

For k = 4

T_4 = (-1)^5 (1/2^4) = -1/16

And so on.

This series can be written as:

1/2, -1/4, 1/8, -1/16, ...

The sum of the terms of this series can be determined by writing out all the 10 terms, and adding them one after the other. That would though, be time consuming.

We can apply the knowledge of Geometric Progression in finding what range the sum is.

From the geometric series, the first term is

a = 1/2

The common ratio is T2/T1 or T3/T2, or ...

r = -1/4 ÷ 1/2 = -1/2

Now, suppose the terms continue infinitely, the sum would be

S = a/(1 - r)

= (1/2)/(1 -(-1/2))

= (1/2)/(3/2)

= 1/3

The series is alternating, we add a number, then subtract a smaller number than that, then add a smaller number, and so on.

This means that the sum of the first 10 terms will very close to 1/3. It may be a little above it or a little below it.

Evidently, it is between 1/4 and 1/2

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step-by-step explanation: