To prove 3+7+11+...+(4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, under our previous assumption, we prove that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. OP(n+1): 3+7+11+. . +4(n+1)-1= 2n² + 5n +31 P(k+1): 3+7+11+...+ 4(k+ 1) − 1 = (k+1)(2(k+1)+1) n+1 P(n+1): (4(+1) − 1) = (n + 1)(2(n + 1) + 1) j=1 n+1 P(n+1): (4j-1) = (n + 1)(2(n+1) + 1) Σw-

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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To prove 3+7+ 11 + ... + (4n − 1) = n(2n + 1) holds for an
arbitrary n € Z+,
in the inductive step, under our previous assumption, we prove that
Check ALL that apply, i.e. you are allowed to choose ONE or MORE
answers.
P(n+1): 3+7+11+.. +4(n+1)-1= 2n² + 5n+3
OP(k+1): 3+7+11+... +4(k+1) − 1 = (k+1)(2(k+1) +1)
n+1
P(n+1): (4(j+1) − 1) = (n + 1)(2(n + 1) + 1)
j=1
n+1
P(n + 1): (4j-1) = (n + 1)(2(n + 1) + 1)
j=1
Transcribed Image Text:To prove 3+7+ 11 + ... + (4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, under our previous assumption, we prove that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. P(n+1): 3+7+11+.. +4(n+1)-1= 2n² + 5n+3 OP(k+1): 3+7+11+... +4(k+1) − 1 = (k+1)(2(k+1) +1) n+1 P(n+1): (4(j+1) − 1) = (n + 1)(2(n + 1) + 1) j=1 n+1 P(n + 1): (4j-1) = (n + 1)(2(n + 1) + 1) j=1
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