2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = ?

1. Translate the problem requirements: We need to find the sum of powers of 2 starting from \(2^1\) and going up to \(2^8\). The expression can be rewritten as \(2^1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\), which simplifies to \(2 \times 2^1 + 2^2 + 2^3 + \ldots + 2^8\).
2. Recognize the geometric series pattern: Identify that we have a sum of consecutive powers of 2, and use the key insight that the sum of powers of 2 from \(2^0\) to \(2^n\) equals \(2^{(n+1)} - 1\).
3. Apply the doubling property strategically: Since our series starts with \(2 \times 2^1\) instead of \(2^0\), adjust our approach by factoring out common terms and using the geometric series sum formula efficiently.
4. Compare with answer choices for verification: Check our calculated result against the given options to confirm we've reached the correct power of 2.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have a sum that begins with 2 + 2, followed by powers of 2 from \(2^2\) up to \(2^8\).

First, let's rewrite the initial part more clearly: \(2 + 2 = 2 \times 2^1 = 2^2\)

So our expression becomes: \(2^2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

This can be simplified to: \(2 \times 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

Which equals: \(2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

Further simplifying: \(2 \times 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 2^4 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

Continuing this pattern: \(2 \times 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 2^5 + 2^5 + 2^6 + 2^7 + 2^8\)

Process Skill: TRANSLATE - Converting the given expression into a more workable form

2. Recognize the geometric series pattern

Now I can see a powerful pattern emerging. Each time I combine two identical powers of 2, I get the next higher power:

• \(2 + 2 = 2^1 + 2^1 = 2^2\)
• \(2^2 + 2^2 = 2^3\)
• \(2^3 + 2^3 = 2^4\)
• And so on...

This tells me something important about sums of consecutive powers of 2. Let me think about this step by step:

If I have \(2^1 + 2^2 + 2^3 + \ldots + 2^8\), there's a beautiful relationship here.

Let me call this sum S. So \(S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

Now, if I double this sum: \(2S = 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^9\)

Subtracting the original from the doubled version:
\(2S - S = 2^9 - 2^1\)
\(S = 2^9 - 2 = 2^9 - 2^1\)

Process Skill: INFER - Drawing the non-obvious conclusion about the doubling relationship

3. Apply the doubling property strategically

But wait! My original expression isn't quite \(2^1 + 2^2 + 2^3 + \ldots + 2^8\).

Let me go back to what I actually have: \(2 + 2 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

This is: \(2^1 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8\)

I can group this as: \((2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8) + 2^1\)

From my work above, I know that \(2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 = 2^9 - 2^1\)

So my total is: \((2^9 - 2^1) + 2^1 = 2^9 - 2 + 2 = 2^9\)

Process Skill: MANIPULATE - Strategically rearranging terms to use the geometric series property

4. Compare with answer choices for verification

My calculated result is \(2^9\).

Looking at the answer choices:
A. \(2^9\) ✓
B. \(2^{10}\)
C. \(2^{16}\)
D. \(2^{35}\)
E. \(2^{37}\)

Let me double-check this makes sense: \(2^9 = 512\)

I can verify with a smaller example: \(2 + 2 + 2^2 = 2 + 2 + 4 = 8 = 2^3\)
Using my method: This would be \(2^1 + 2^1 + 2^2 = (2^1 + 2^2) + 2^1 = (2^3 - 2^1) + 2^1 = 2^3\) ✓

Final Answer

The answer is A. \(2^9\)

The key insight was recognizing that the sum of consecutive powers of 2 follows the pattern where the sum from \(2^1\) to \(2^n\) equals \(2^{(n+1)} - 2^1\), and then adjusting for the extra \(2^1\) term in our original expression.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the initial terms: Students often fail to recognize that "2 + 2" at the beginning should be rewritten as "\(2^1 + 2^1\)" to fit the pattern. They might try to work with the expression as given without standardizing the notation, making it harder to see the geometric series structure.

2. Not recognizing the geometric series pattern: Many students will attempt to calculate each power of 2 individually (2=2, 4=4, 8=8, etc.) and then add them up manually, rather than recognizing this as a geometric series that can be solved using the formula for sum of geometric series.

3. Confusion about which geometric series formula to apply: Even when students recognize it's a geometric series, they might struggle with whether to use the standard formula \(S = a(r^n-1)/(r-1)\) directly or use the doubling method shown in the solution, leading to setup errors.

Errors while executing the approach

1. Arithmetic errors in the doubling method: When applying \(2S - S = 2^9 - 2^1\), students frequently make calculation mistakes, such as forgetting that \(2^9 - 2^1 = 2^9 - 2\) (not \(2^9 - 1\)), or incorrectly handling the subtraction of terms.

2. Incorrect handling of the extra \(2^1\) term: Students often get confused when accounting for the fact that the original expression has an extra "2" (or \(2^1\)) compared to the standard geometric series. They might add it incorrectly or forget to include it in their final calculation.

3. Powers of 2 calculation errors: Basic computational mistakes when working with exponents, such as miscalculating what \(2^9\) equals or confusing the relationship between consecutive powers of 2.

Errors while selecting the answer

1. Confusing the final exponent: After correctly calculating that the sum equals \(2^9\), students might second-guess themselves and select \(2^{10}\) (thinking they need "one more" power) or make other exponent-related errors when matching their result to the answer choices.

2. Not verifying the reasonableness of the answer: Students might arrive at an answer but fail to do a quick sanity check. For instance, they could select extremely large values like \(2^{35}\) or \(2^{37}\) without realizing these numbers are far too large for a sum that starts with small terms like 2, 2, 4, 8.