Question

Prove the following theorem by mathematical induction: Theorem: the sum of the first n even integers is n(n 1), i.e., 2+ 4 +... +2n=n(n+ 1) Proof (by induction): let P(k) represent the proposition: P(k): 2+ 4+ ... +2k=k(k+ 1) a) Base case: (n=1): show that the base case P(1) is true b) inductive step: prove P(k) --->P(k +1) (i.e., if P(k) is true, then P(k+ 1) is true):

Name: prove the following theorem by mathematical induction theorem the sum of the first n even integers is nn 1 ie 2 4 2nnn 1 proof by induction let pk represent the proposition pk 2 4 2kkk 1 a b 39358
Uploaded: 2024-01-17T20:55:41-08:00
Duration: 3 min 1 s
Channel: Gregory Higby
Description: prove the following theorem by mathematical induction theorem the sum of the first n even integers is nn 1 ie 2 4 2nnn 1 proof by induction let pk represent the proposition pk 2 4 2kkk 1 a b 39358

          Prove the following theorem by mathematical induction: 
Theorem: the sum of the first n even integers is n(n 1), i.e., 
2+ 4 +... +2n=n(n+ 1)
Proof (by induction): let P(k) represent the proposition: 
P(k): 2+ 4+ ... +2k=k(k+ 1)
a) Base case: (n=1): show that the base case P(1) is true
b) inductive step: prove P(k) --->P(k +1) (i.e., if P(k) is true, then P(k+ 1) is true):

Added by Guillermo H.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Gregory Higby

01/17/2024

Step 1

P(1) states that the sum of the first 1 even integer is equal to 1(1 + 1). Calculating the left side: 2 = 2. Calculating the right side: 1(1 + 1) = 1(2) = 2. Since both sides are equal, P(1) is true. Show more…

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Prove the following theorem by mathematical induction: Theorem: the sum of the first n even integers is n(n 1), i.e., 2+ 4 +... +2n=n(n+ 1) Proof (by induction): let P(k) represent the proposition: P(k): 2+ 4+ ... +2k=k(k+ 1) a) Base case: (n=1): show that the base case P(1) is true b) inductive step: prove P(k) --->P(k +1) (i.e., if P(k) is true, then P(k+ 1) is true):