Prove that 7^n - 6n -1 is divisible by 36 for any natural number, n. [Mathematical Induction]

[@ayaanmuslimcareafrica6623]

the best way ever to understand Mathematical induction

[@jensberling2341]

The love able and lovely performance, and the subtle and clear persuasion of this math. Induction is breathtaking. Yes, breathtaking for me who has struggled to understand it. Thank you, Doctor

[@flowersArePretty1]

YOU ARE GENIUS MAN! THANK YOU, this has helped so much !

[@EzraSchroeder]

this one is straight out of an advanced calculus book haha! it is a problem in chapter one of Kenneth A. Ross: Analysis the Theory of Calculus -- love it!

[@benjaminvatovez8823]

Thank you very much for your video. I would have suggested to use the binomial theorem:
7^n=(6+1)^n=1+6n+36(sum with k from 2 to n) C^k_n 6^(k-2)

[@giuliofalco9816]

Using congruence i proved in this way: 36 = 2^2 3^2 so I apply chinese theorem of remainder writing the following congruence systems: 7^n - 6n - 1 = 0 mod 2; and 7^n - 6n - 1 = 0 mod 3. Substitute 7 with and 6 with the remainder of nteger division by 2 and 3 and the equivalence follow.

[@lush1523]

Am gonna be here till I finish my degree, I have a discrete mathematics course this semester

[@KamalAzhar-t7q]

To prove such property :> (where f is a given fonction and p is a fixed integer) one generally can use induction. But the following is much harder

[@robertveith6383]

Or, look at
(6 + 1)^n - 6n - 1 =
6^n + n*6^(n - 1)*1^1 + [n(n - 1)/2]*6^(n - 2)*1^2 + ... + [n(n - 2)/2]*6^2*1^(n - 2) + n*6^1*1^(n - 1) +
1^n - 6n - 1 =
6^n + n*6^(n - 1) + ... + [n(n - 2)/2]*36 + 6n + 1 - 6n - 1 =
6^n + n*6^(n - 1) + ... + [n(n - 2)/2]*36
That is divisible by 36.

[@flowersArePretty1]

Best teacher

[@gurdiprooprai6453]

Excellent work, but the most difficult part of this video is what determines whether or not you cross your "7's" or not. I am still working on it lol.

[@BP-gn2cl]

Write 7^n =(6+1) ^n, apply bionomial expansion and in fourth line you can prove the required thing

[@arthurkassis]

Take the first possibility:
n=1, 7-6-1=0, which is divisible by 36
then, assume that it's true for n=k, (k belongs to naturals)
7ˆk-6k-1=36m, (m belongs to naturals)
now, let's see if it also works when n=k+1
7ˆ(k+1)-6(k+1)-1
=7.7ˆk-6k-6-1
=7.7ˆk-6k-7
=7(7ˆk-1)-6k
=7(36m+6k)-6k
=7.36.m+42k-6k
=7.36m+36k, which is divisible by 36 as both terms are multiples of it.

[@ruthgonzalez6547]

Nice!

[@sumeetmahapatra9862]

Thank you sir

[@zazavitch1]

❤

[@aayushixi]

omg tysmmmm

[@abdussamadabubakar1746]

Damn!!!!!