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Discrete Math. Prove that the sum of squares of the first n positive integers = (n(n+1)(n+2))/6
In this video, we prove by induction that the sum of square of the first n positive integers is equal to (n(n+1)(n+2))/6. This problem was taken from Discrete Mathematics and Its Applications by Kenneth Rosen, 7th edition, Chapter 5.1, question 3.
1^2 + 2^2 + 3^2 + ... + n^2 = (n(n+1)(n+2))/6
Discrete Mathematics playlist:
https://youtube.com/playlist?list=PLm90IN9RVLf_BneWC40564ZZAqpe2sz6-&si=bKhYao84EXCHpl6N
Induction Proofs playlist:
https://youtube.com/playlist?list=PLm90IN9RVLf-z-V3NIPi0-ZhxckZHup9q&si=hIv_gDttX16fM1F5
Chapters:
00:00 Introduction to Question
01:13 Base Case P(1)
02:58 Inductive Step
03:20 Induction Hypothesis P(k)
04:51 We Want to Show P(k+1) Case
06:53 Algebra Steps
12:23 QED Thanks for Watching
#discrete_mathematics #discretemathematics #induction_proof #induction #integers #integerproof #sumofcubes
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