Discrete Math II - 5.1.1 Proof by Mathematical Induction

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Kimberly Brehm

Kimberly Brehm

Күн бұрын

Пікірлер: 13
@JosephCote-y3j
@JosephCote-y3j 8 ай бұрын
You should take the sum of I.H (k^2) and substitute that in for 1 + 3 + 5 + … (2k - 1) since we are assuming they equal each other. So you can now do, k^2 + 2k + 1 = (k + 1)^2.
@artemkrazhan8603
@artemkrazhan8603 Ай бұрын
ur the goat prof kimberly thank you so much for this video
@karqoa3968
@karqoa3968 2 жыл бұрын
thank you very very much for these videos
@Userpng77y
@Userpng77y Жыл бұрын
Thanks for the course, but this should be seen according to the roden of the youtube playlist or according to the number of the videos "5.1.1, 5.1.2......"?
@XxNGameCubexX
@XxNGameCubexX 23 күн бұрын
Slight mistake on the well-ordering principle slide, I think the professor meant to say S is a subset of Z+, Q+, or R+, respectively, not element of. Thanks for your lectures!
@AdrianMeyer8
@AdrianMeyer8 12 күн бұрын
Very appreciative for making the making these courses available, Kimerbly! But - however I am a bit confused, how does 1 + 3 + 5 + ... + (2k - 1) + (2k +1) = (k + 1)^2? I understand if you would swap "1 + 3 + 5 + ... + (2k - 1)" with the "I.H" witch states "k^2". But teacher in the video doesn't explain this. It would be a lot more time saving if she would adress that when writing 1 + 3 + 5 + ... + (2k - 1) is just the I.H and the I.H is k^2 and next step is to add (2k+1).
@SawFinMath
@SawFinMath 11 күн бұрын
I'll take another look at that video. Thanks!
@sedibe
@sedibe Жыл бұрын
you r the best
@alifnm3607
@alifnm3607 Жыл бұрын
@caitbenn365
@caitbenn365 Жыл бұрын
Great video. Why did you add 2k+1 to both sides?
@SawFinMath
@SawFinMath Жыл бұрын
That is the definition of an odd integer
@feltonkaonga
@feltonkaonga 2 жыл бұрын
❤👍
@kario360
@kario360 Жыл бұрын
very complex video
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