20 j ΣΣ (2k + 3) = j=5 k=0 = 2n+3 Σ (n + 5k)3 k=0 first step 2nd step final answer first step and
20 j ΣΣ (2k + 3) = j=5 k=0 = 2n+3 Σ (n + 5k)3 k=0 first step 2nd step final answer first step and
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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Show the first few steps and the final answer in evaluating the given summation. Final answers must be in its simplest whole or rational number, or expression.
![20 j
ΣΣ (2k + 3) =
j=5 k=0
2n+3
Σ (n + 5k) =
k=0
||
first step
2nd step
final answer
first step
2nd step
final answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5260fad0-d07c-4182-bd39-9605ad362b03%2Fa1c05619-23b2-4735-9b23-d88417e97b52%2F9ws0hpg_processed.png&w=3840&q=75)
Transcribed Image Text:20 j
ΣΣ (2k + 3) =
j=5 k=0
2n+3
Σ (n + 5k) =
k=0
||
first step
2nd step
final answer
first step
2nd step
final answer
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