For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by \((-1)^{(\mathrm{k}+1)}...

1. Translate the problem requirements: We need to find the sum T of the first 10 terms of a sequence where each k-th term equals \((-1)^{(k+1)} \times \frac{1}{2^k}\). This creates an alternating series with terms that get progressively smaller.
2. Generate the actual terms to identify the pattern: Calculate the first few terms to see the alternating positive/negative pattern and how the magnitudes decrease.
3. Group terms strategically to simplify calculation: Since we have an alternating series, pair consecutive terms to see how they partially cancel out, making the sum easier to evaluate.
4. Estimate the final sum range: Use the grouped terms and remaining terms to determine which answer choice range contains our sum T.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have a sequence where each term follows a specific pattern, and we need to find the sum of the first 10 terms.

The formula given is \((-1)^{(k+1)} \times \frac{1}{2^k}\) for the k-th term. Let's understand what this means in plain English:

• The \((-1)^{(k+1)}\) part creates an alternating pattern of positive and negative signs
• The \(\frac{1}{2^k}\) part means we're dividing by 2 raised to the k-th power, so terms get smaller and smaller
• We need to add up the first 10 of these terms to get T

Process Skill: TRANSLATE - Converting the mathematical notation into understandable language

2. Generate the actual terms to identify the pattern

Let's calculate the first several terms to see the pattern clearly:

For k=1: \((-1)^{(1+1)} \times \frac{1}{2^1} = (-1)^2 \times \frac{1}{2} = 1 \times \frac{1}{2} = \frac{1}{2}\)
For k=2: \((-1)^{(2+1)} \times \frac{1}{2^2} = (-1)^3 \times \frac{1}{4} = -1 \times \frac{1}{4} = -\frac{1}{4}\)
For k=3: \((-1)^{(3+1)} \times \frac{1}{2^3} = (-1)^4 \times \frac{1}{8} = 1 \times \frac{1}{8} = \frac{1}{8}\)
For k=4: \((-1)^{(4+1)} \times \frac{1}{2^4} = (-1)^5 \times \frac{1}{16} = -1 \times \frac{1}{16} = -\frac{1}{16}\)
For k=5: \((-1)^{(5+1)} \times \frac{1}{2^5} = (-1)^6 \times \frac{1}{32} = 1 \times \frac{1}{32} = \frac{1}{32}\)
For k=6: \((-1)^{(6+1)} \times \frac{1}{2^6} = (-1)^7 \times \frac{1}{64} = -1 \times \frac{1}{64} = -\frac{1}{64}\)

Continuing this pattern:
k=7: \(\frac{1}{128}\), k=8: \(-\frac{1}{256}\), k=9: \(\frac{1}{512}\), k=10: \(-\frac{1}{1024}\)

So our sequence starts: \(\frac{1}{2}, -\frac{1}{4}, \frac{1}{8}, -\frac{1}{16}, \frac{1}{32}, -\frac{1}{64}, \frac{1}{128}, -\frac{1}{256}, \frac{1}{512}, -\frac{1}{1024}\)

3. Group terms strategically to simplify calculation

Since we have an alternating series, let's pair consecutive terms to see how they partially cancel:

Pair 1: \(\frac{1}{2} + \left(-\frac{1}{4}\right) = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}\)
Pair 2: \(\frac{1}{8} + \left(-\frac{1}{16}\right) = \frac{2}{16} - \frac{1}{16} = \frac{1}{16}\)
Pair 3: \(\frac{1}{32} + \left(-\frac{1}{64}\right) = \frac{2}{64} - \frac{1}{64} = \frac{1}{64}\)
Pair 4: \(\frac{1}{128} + \left(-\frac{1}{256}\right) = \frac{2}{256} - \frac{1}{256} = \frac{1}{256}\)
Pair 5: \(\frac{1}{512} + \left(-\frac{1}{1024}\right) = \frac{2}{1024} - \frac{1}{1024} = \frac{1}{1024}\)

Notice the beautiful pattern! Each pair sums to exactly half of the first term in that pair.

So \(T = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\)

Process Skill: SIMPLIFY - Strategic pairing makes the calculation much more manageable

4. Estimate the final sum range

Now let's add up our paired terms: \(T = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\)

Let's convert everything to a common denominator. Since \(1024 = 4 \times 256 = 16 \times 64 = 64 \times 16 = 256 \times 4\), we can use 1024:

\(\frac{1}{4} = \frac{256}{1024}\)
\(\frac{1}{16} = \frac{64}{1024}\)
\(\frac{1}{64} = \frac{16}{1024}\)
\(\frac{1}{256} = \frac{4}{1024}\)
\(\frac{1}{1024} = \frac{1}{1024}\)

\(T = \frac{256 + 64 + 16 + 4 + 1}{1024} = \frac{341}{1024}\)

To check which range this falls into:

• \(\frac{1}{4} = \frac{256}{1024}\)
• \(\frac{1}{2} = \frac{512}{1024}\)

Since \(256 < 341 < 512\), we have \(\frac{1}{4} < T < \frac{1}{2}\)

To be more precise: \(\frac{341}{1024} \approx 0.333\), which is indeed between \(\frac{1}{4} = 0.25\) and \(\frac{1}{2} = 0.5\)

Final Answer

\(T = \frac{341}{1024}\), which is approximately 0.333. This value lies between \(\frac{1}{4}\) (0.25) and \(\frac{1}{2}\) (0.5).

Therefore, the answer is D. Between 1/4 and 1/2.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the alternating sign pattern

Students often confuse the exponent in \((-1)^{(k+1)}\) and incorrectly assume the sequence starts with a negative term. They might think when k=1, \((-1)^{(k+1)} = (-1)^2 = -1\) instead of +1, leading to an entirely wrong sequence pattern. This is critical because getting the sign pattern wrong affects every subsequent calculation.

2. Attempting to use infinite geometric series formula inappropriately

Seeing the terms like \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}\), students might immediately think this is a standard geometric series and try to apply the formula \(S = \frac{a}{1-r}\). However, this approach fails because: (a) the series alternates signs, and (b) we only need the first 10 terms, not the infinite sum. This misconception leads students down the wrong solution path entirely.

Errors while executing the approach

1. Arithmetic errors when pairing terms

When calculating pairs like \(\frac{1}{2} + \left(-\frac{1}{4}\right)\), students often make basic fraction arithmetic mistakes. For example, they might incorrectly compute \(\frac{1}{2} - \frac{1}{4}\) as \(\frac{1-1}{2-4} = \frac{0}{-2} = 0\), instead of finding the common denominator: \(\frac{2}{4} - \frac{1}{4} = \frac{1}{4}\). These errors compound throughout the calculation.

2. Errors in converting fractions to common denominators

When adding the final sum \(T = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\), students frequently make mistakes in converting to the common denominator of 1024. For instance, they might incorrectly convert \(\frac{1}{4}\) to \(\frac{254}{1024}\) instead of \(\frac{256}{1024}\), or make similar errors with other fractions, leading to an incorrect final sum.

Errors while selecting the answer

1. Incorrectly comparing the final result to the given ranges

After correctly calculating \(T = \frac{341}{1024}\), students may struggle with comparing this fraction to the boundary values \(\frac{1}{4}\) and \(\frac{1}{2}\). They might incorrectly convert these to decimals (getting confused about whether \(\frac{341}{1024}\) is approximately 0.33 or 0.034) or make errors when converting \(\frac{1}{4}\) and \(\frac{1}{2}\) to the denominator 1024 for comparison, leading them to select the wrong range.