Question

Find the sum of series: \(2*2^2 - 2*3^2 + 2*4^2 +\) \(….(20 \,terms)\)

• A. 462
• B. -462
• C. 460
• D. -460

Answer 1

The Correct Option is D

Solution:

First, notice that the pattern appears to alternate between a coefficient that is either \(2\) or an odd integer (\(3,5,7,\dots\)) multiplied by a power of \(2^n\) or \((2n)^2\). However, the provided solution snippet shows a regrouping approach:

\[ (\underbrace{2^2 - 3\cdot 2^3 + 4^2 - 5\cdot 2^5 + \dots + 20\text{ terms}}_{\text{Group A}}) + (\underbrace{2^2 + 4^2 + \dots + (2\cdot 10)^2}_{\text{10 even squares}}) \]

\[ - (\underbrace{2 + 3 + 4 + 5 + \dots + 11}_{\text{some consecutive integers}}) + 4\Bigl[\underbrace{1^2 + 2^2 + \dots + 10^2}_{\text{sum of squares up to }10}\Bigr]. \]

The snippet then shows evaluating several known sums:

• Sum of first \(n\) integers: \(\frac{n(n+1)}{2}\).
• Sum of first \(n\) squares: \(\frac{n(n+1)(2n+1)}{6}\).

After carefully combining like terms (and possibly correcting sign patterns), the final numeric answer shown is:

\[ 1310. \]

Therefore, the sum of the first 20 terms of the given series is \(\boxed{1310}\).

Analysis:

The series is: \( 2^2 - 3 \cdot 2^3 + 4^2 - 5 \cdot 2^5 + 6^2 - 7 \cdot 2^7 + \cdots \)

The solution snippet proposes a regrouping strategy:

• Even Squares: \( 2^2 + 4^2 + \cdots + 20^2 = 4(1^2 + 2^2 + \cdots + 10^2) \)
• Consecutive Integers: \( 2 + 3 + 4 + \cdots + 11 \)
• Group A: \( 2^2 - 3 \cdot 2^3 + 4^2 - 5 \cdot 2^5 + \cdots \) (This is the crucial part that's not explicitly calculated.)

Using Sum Formulas (Correctly Applied)

• Sum of Squares: \( 4 \sum_{k=1}^{10} k^2 = 4 \cdot \frac{10(11)(21)}{6} = 1540 \)
• Sum of Integers: \( \sum_{k=2}^{11} k = \frac{(2+11)(10)}{2} = 65 \)

The Missing Calculation: Group A

The key issue is that the snippet doesn't show how to calculate Group A. This group consists of alternating terms with increasing powers of 2 and odd coefficients.

Why This Is Important

Without calculating Group A, we can't verify the final answer of 1310. It's essential to understand how the alternating terms contribute to the overall sum.

What We Know (and Don't Know)

• We know how the even squares and consecutive integers are calculated.
• We know the final answer is stated as 1310.
• We don't know the exact calculation for Group A.

The Need for a Complete Solution

To provide a fully verified solution, we would need to:

1. Derive a general form for the terms in Group A.
2. Find a method to sum the alternating series in Group A.
3. Combine the result from Group A with the sums of even squares and consecutive integers.

In conclusion: While the snippet provides a useful regrouping strategy and correctly applies sum formulas, it lacks the crucial calculation of Group A, which is necessary to confirm the final answer of 1310.