I understand the reasoning and the logic behind every step you take. I just don't get how you come up with things like (k-1) * k * (k+1) and think of them as consecutive numbers, where atleast one will be even, and another a multiple of 3. Maybe I just don't think in the right way, but I can't come up with these answers myself, though it makes a lot of sense when you go through them.
for the second example: so we dont have to worry about the k>1 in our working?? cos some questions need you to think about it to get your answer and others dont, so im just confused on when i do or dont need to look at and worry about it. Great video by the way, thankyouu
k has been defined as greater than 1 in the conjecture, we're told to prove it for those values. It's not that we're saying the statement is not true for other values of k also.
Square brackets in the titles tells you that it is also AS Maths material. When you learn a topic in 1st / 2nd year will depend on the school / college
Hi sir with the second example I set the k as (n^2+1) because it has to be greater than 1 and also positive but when I did the full expansion I got n^6+ 3n^4+ 2n^2. Have I expanded it wrong our is there a way to show that that's divisible by 6?
n^6+ 3n^4+ 2n^2 is correct, but you haven't done yourself any favours. It can be factorised to get n^2(n^4 + 3n^2 + 2) = n^2(n^2 + 1)(n^2 + 2) This is precisely the same as factorising k^3 - k = k(k^2 - 1) = k(k-1)(k+1) and then substituting k = n^2 + 1. Either way, you'll still have to go through the explanation as shown in the video.
Can you help me with this question, Prove that the statement is true for all natural number using mathematical induction, when (1 ' 2) + (2 ' 3) + (3 ' 4) + ...... + n(n+1) = 1/3 n(n+1)(n+2)
If you have 8 - 3 - 5, this is equal to 0. It is not the same as 8 - (3 - 5) which would make 8 - 3 + 5 So n^2 + 5n + 6 - n^2 - n = 5n + 6 - n = 4n + 6
Third one? Do you mean the second example? There's no much else you can do with k^3 - k apart from factorise it, then you would need to spot that it is the product of three consecutive integers.
can you help me with this, given that un=1 , un=2 and un= un-1 + un-2 (n>2), use the principle of strong mathematical induction to prove that un< (7/4)^n , n greater than or equal to 1.
yes, and this is another brilliant example of consequences and equivalences isn't it. "Product of three consecutive integers gets an integer divisible by 6" ; this however only works in this direction. If one considers the converse, it would be: "an integer divisible by 6 is made of product of three consecutive integers", which is obviously false statement)
![A-Level Maths: A1-08 [Introducing Disproof by Counter Example]](http://i.ytimg.com/vi/Q9esB3ZSAPc/mqdefault.jpg)
![A-Level Maths: A1-08 [Introducing Disproof by Counter Example]](/img/tr.png)
This teacher is so good. Honestly your a great help man.
Thank you!
Jack Brown keep up the work
I understand the reasoning and the logic behind every step you take. I just don't get how you come up with things like (k-1) * k * (k+1) and think of them as consecutive numbers, where atleast one will be even, and another a multiple of 3. Maybe I just don't think in the right way, but I can't come up with these answers myself, though it makes a lot of sense when you go through them.
ikr
You are seriously an amazing teacher. Each day my confidence in maths improves by watching your videos! Many thanks :)
ur actually the best maths youtuber out there
I thought I would never understand this topic but this video helped a lot, thank you
Thank you from 2019. Why wouldn't someone not give in Algebraic explanation, whenever the mathematical problem isn't simple?
Thank you for the videos they're great.
Thank you this was a great video.
I have discovered great new channel
It really helped. Thanks a lot.
Very clear explaination and easy to understand even a difficult questions
Exceptional teaching
I'm a little confused to how you got the 5n when expanding the brackets. I have clearly missed something. Could you please explain if possible :)
(n+2)*(n+3) = n^2 + 2n + 3n + 6 = n^2 + 5n + 6
Expand using a grid if necessary.
@@TLMaths Thank you :) I understand it now! You're awesome
for the second example: so we dont have to worry about the k>1 in our working?? cos some questions need you to think about it to get your answer and others dont, so im just confused on when i do or dont need to look at and worry about it. Great video by the way, thankyouu
In this example it's just there so we don't need to worry about k=0 and negative k. But yes, you should always think how it might play in the question
Very good and excellent teaching 👍
Jack thanks found Q2 helpful... was wondering why if we try proof by exhaustion does not work with k equals 2n+1 and 2n
i love your vids so much
thank you so much, I really needed this
Solid video
I would leave a like
Example 2, why k>1 as k=0 or k=1 won't contradict the conjecture?
k has been defined as greater than 1 in the conjecture, we're told to prove it for those values. It's not that we're saying the statement is not true for other values of k also.
agree and raised it in case that becomes a question of someone else in the future. Thanks Jack for valuable videos by the way!
for 2nd example, why must it be divisible by 2? if there are 2 evens then it wont be? (eg 4+5+6 = 15)
*Product* of three consecutive integers (4x5x6), *not* the sum.
For the first one is it possible to use 4n-4 4n-3 4n-2 4n-1?
For some reason it didn’t work when I did it, but when I flipped the signs it did. Does anyone know why this is the case?
(4n-1)(4n-2) - (4n-3)(4n-4) = 16n - 10
(4n-1) + (4n-2) + (4n-3) + (4n-4) = 16n - 10
So it DOES work for these.
Where does the year one content end and the year two content start
Square brackets in the titles tells you that it is also AS Maths material. When you learn a topic in 1st / 2nd year will depend on the school / college
You know for the second example. Cant you use a value greater than 1 for K and prove it this way?
You would have to check it works for every value of k > 1. One example cannot prove the statement.
Thank you so much !!!
Can we take n ,2n ,2n+1,2n+2???
n and 2n aren't consecutive integers unless n=1. You could try 2n, 2n+1, 2n+2, 2n+3 (although I'm not sure why you would!)
(2n+2)(2n+3) - (2n)(2n+1)
= (4n^2+10n+6) - (4n^2 + 2n)
= 8n + 6
(2n) + (2n+1) + (2n+2) + (2n+3)
= 8n + 6
Hi sir with the second example I set the k as (n^2+1) because it has to be greater than 1 and also positive but when I did the full expansion I got n^6+ 3n^4+ 2n^2. Have I expanded it wrong our is there a way to show that that's divisible by 6?
n^6+ 3n^4+ 2n^2 is correct, but you haven't done yourself any favours.
It can be factorised to get n^2(n^4 + 3n^2 + 2) = n^2(n^2 + 1)(n^2 + 2)
This is precisely the same as factorising k^3 - k = k(k^2 - 1) = k(k-1)(k+1) and then substituting k = n^2 + 1.
Either way, you'll still have to go through the explanation as shown in the video.
I know this was 2 years ago but can you even set k as n^2 + 1 because if n=0 then k=1 and k has to be greater than 1
Great video!
Can you help me with this question, Prove that the statement is true for all natural number using mathematical induction, when (1 ' 2) + (2 ' 3) + (3 ' 4) + ...... + n(n+1) = 1/3 n(n+1)(n+2)
Second starred example: www.purplemath.com/modules/inductn3.htm
thanks
how ,in the first one, is 5n - -n = 4n? wouldnt it be 6n?
If you have 8 - 3 - 5, this is equal to 0. It is not the same as 8 - (3 - 5) which would make 8 - 3 + 5
So n^2 + 5n + 6 - n^2 - n
= 5n + 6 - n
= 4n + 6
how is one supposed to just predict the direction of the third one..
Third one? Do you mean the second example?
There's no much else you can do with k^3 - k apart from factorise it, then you would need to spot that it is the product of three consecutive integers.
word i factorised but that is a stretch to "just see" it
Thanks, I understand the first one, but the second question went over my head, gonna rewatch it but if you can help me Jack, I would appreciate it.
Let me know if there’s any particular part of the proof you want me to help you with
can you help me with this, given that un=1 , un=2 and un= un-1 + un-2 (n>2), use the principle of strong mathematical induction to prove that un< (7/4)^n , n greater than or equal to 1.
I don’t know the ins and outs of “strong” mathematical induction I’m afraid.
can you do a level maths
sometimes
im studying that and want to go on to a level further maths any chance of vids for all the topics
I've done A-Level Maths in this playlist. I haven't done Further
thanks that the one i want to get up to minimum
You can navigate all the videos I've done more easily using my website:
sites.google.com/site/tlmaths314/home/a-level-maths-2017
How about 2 4 and 5 non of them are divisible by 3 ?
245 are not consecutive integers though
yes, and this is another brilliant example of consequences and equivalences isn't it. "Product of three consecutive integers gets an integer divisible by 6" ; this however only works in this direction.
If one considers the converse, it would be: "an integer divisible by 6 is made of product of three consecutive integers", which is obviously false statement)
i learnt number 2 in gcse wow