Compute the summation. (Enter an exact number.) 3 Σ 4k2 + 6 k = 1 II
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Summation Problem**
Compute the summation. (Enter an exact number.)
\[ \sum_{k=1}^{3} \left( 4k^2 + 6 \right) = \]
Explanation:
The given mathematical problem involves computing the summation of the expression \(4k^2 + 6\) as the variable \(k\) ranges from \(1\) to \(3\).
This means you will need to:
1. Substitute \( k = 1 \) into the expression and calculate the result.
2. Substitute \( k = 2 \) into the expression and calculate the result.
3. Substitute \( k = 3 \) into the expression and calculate the result.
Finally, you will sum all these individual results to find the final answer.
Here are the steps in detail:
1. For \( k = 1 \):
\[
4(1)^2 + 6 = 4 \times 1 + 6 = 10
\]
2. For \( k = 2 \):
\[
4(2)^2 + 6 = 4 \times 4 + 6 = 22
\]
3. For \( k = 3 \):
\[
4(3)^2 + 6 = 4 \times 9 + 6 = 42
\]
Adding all these results:
\[
10 + 22 + 42 = 74
\]
Therefore, the sum is:
\[
74
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2b4e390-596f-48e2-849d-32f83635f844%2F4e021e42-769c-45eb-b440-fcdb2d4db398%2Fu1d7flk.png&w=3840&q=75)
Transcribed Image Text:**Summation Problem**
Compute the summation. (Enter an exact number.)
\[ \sum_{k=1}^{3} \left( 4k^2 + 6 \right) = \]
Explanation:
The given mathematical problem involves computing the summation of the expression \(4k^2 + 6\) as the variable \(k\) ranges from \(1\) to \(3\).
This means you will need to:
1. Substitute \( k = 1 \) into the expression and calculate the result.
2. Substitute \( k = 2 \) into the expression and calculate the result.
3. Substitute \( k = 3 \) into the expression and calculate the result.
Finally, you will sum all these individual results to find the final answer.
Here are the steps in detail:
1. For \( k = 1 \):
\[
4(1)^2 + 6 = 4 \times 1 + 6 = 10
\]
2. For \( k = 2 \):
\[
4(2)^2 + 6 = 4 \times 4 + 6 = 22
\]
3. For \( k = 3 \):
\[
4(3)^2 + 6 = 4 \times 9 + 6 = 42
\]
Adding all these results:
\[
10 + 22 + 42 = 74
\]
Therefore, the sum is:
\[
74
\]
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