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The inductive step of an inductive proof shows that for k≥1, if Σ½-1 j(j + 1) = \/ k(k+ 1)(k + 2), then k+1 Σ1 j(j + 1) = (k+ 1)(k+ 2)(k+ 3) Parts of the proof are shown below. Select the expression that should replace the [?]. k+1 j(j+1) j=1 k+1 j(j+1) j=1 k+1 -j=1 j(j + 1) k+1 = [?] (Step 1) = · ¼k(k + 1)(k + 2) + (k+1) (k+2) By the inductive hypothesis (Algebraic Steps) j(j+1) = (k+ 1) (k + 2)(k + 3) k+1 ο Σ#1 3(j + 1) + k(k + 1) j=1 ο Σ1 3(j + 1) + (k + 1)(k + 2) ○ Σ -1 j(j + 1) + k(k+1) ΟΣ+1 jj + 1) + (k + 1)(k + 2) j=1

The inductive step of an inductive proof shows that for k≥1, if Σ½-1 j(j + 1) = \/ k(k+ 1)(k + 2), then k+1 Σ1 j(j + 1) = (k+ 1)(k+ 2)(k+ 3) Parts of the proof are shown below. Select the expression that should replace the [?]. k+1 j(j+1) j=1 k+1 j(j+1) j=1 k+1 -j=1 j(j + 1) k+1 = [?] (Step 1) = · ¼k(k + 1)(k + 2) + (k+1) (k+2) By the inductive hypothesis (Algebraic Steps) j(j+1) = (k+ 1) (k + 2)(k + 3) k+1 ο Σ#1 3(j + 1) + k(k + 1) j=1 ο Σ1 3(j + 1) + (k + 1)(k + 2) ○ Σ -1 j(j + 1) + k(k+1) ΟΣ+1 jj + 1) + (k + 1)(k + 2) j=1

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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The inductive step of an inductive proof shows that for k≥1, if Σ½-₁ j(j + 1) = \/ k(k+ 1)(k + 2), then k+1 Σ1 j(j + 1) = (k+ 1)(k+ 2)(k+ 3) Parts of the proof are shown below. Select the expression that should replace the [?]. k+1 j(j+1) {j(j + 1) k+1 j=1 k+1 -j=1 j(j+1) k+1 j(j+1) = (Step 1) · ¼k(k + 1)(k + 2) + (k+1) (k+2) By the inductive hypothesis (Algebraic Steps) [?] =... = = (k+ 1) (k+ 2) (k+ 3) ○ k+1 j=1 j(j + 1) + k(k+1) ο Σ1 3(j + 1) + (k + 1)(k + 2) ○ Σ -1 j(j + 1) + k(k+1) =1 ΟΣ+1 jj + 1) + (k + 1)(k + 2)
Transcribed Image Text:The inductive step of an inductive proof shows that for k≥1, if Σ½-₁ j(j + 1) = \/ k(k+ 1)(k + 2), then k+1 Σ1 j(j + 1) = (k+ 1)(k+ 2)(k+ 3) Parts of the proof are shown below. Select the expression that should replace the [?]. k+1 j(j+1) {j(j + 1) k+1 j=1 k+1 -j=1 j(j+1) k+1 j(j+1) = (Step 1) · ¼k(k + 1)(k + 2) + (k+1) (k+2) By the inductive hypothesis (Algebraic Steps) [?] =... = = (k+ 1) (k+ 2) (k+ 3) ○ k+1 j=1 j(j + 1) + k(k+1) ο Σ1 3(j + 1) + (k + 1)(k + 2) ○ Σ -1 j(j + 1) + k(k+1) =1 ΟΣ+1 jj + 1) + (k + 1)(k + 2)
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