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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
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
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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