Induction Help! (1 Viewer)

[lookoutastroboy]

Hey guys,

I'm trying to do the 1999 Q5 induction but I have no idea what is happening. Can anyone kindly explain the result? - especially the (k+1)th term

Page 67 on the doc.





Cheers,
l.a.

 

[hscishard]

LHS= (k+2)(k+3)...(2(k+1)-1)(2k+2)
=(k+2)...(2k+1)(2k+2)
=2^n [1*3*...(2k-1)] * (2k+1)(2k+2)/ (k+1)
=2^(n+1) [1*3*...(2k+1)]

=RHS(cbb doing this part, but it's really simple)

Line 3. It's all over k+1 because the expression for n=k+1 doesnt begin with k+1, but the assumption uses the k+1. Hence diving it will leave the thing unaltered

 

[Trebla]

The question is to prove by induction that for positive integers n



Proving for n = 1 is trivial, now assume the statement holds for n = k



We need to prove that



If the statement holds for n = k then it holds for n = k + 1. Since it holds for n = 1, then it holds for all positive integers n by induction.

 

[lookoutastroboy]

The question is to prove by induction that for positive integers n



Proving for n = 1 is trivial, now assume the statement holds for n = k



We need to prove that



If the statement holds for n = k then it holds for n = k + 1. Since it holds for n = 1, then it holds for all positive integers n by induction.

Thanks guys - eh trebla how did you get the 2nd line of the LHS side? I also don't get the (k+1)th term

 

[funnytomato]

Thanks guys - eh trebla how did you get the 2nd line of the LHS side? I also don't get the (k+1)th term

Have you done it at school ? If not, read the relevant section of the textbook and AFTER you've learnt the concept, do the questions.

 

[funnytomato]

basically there are 4 steps

1. prove result is true for some initial value, usually n=1
2 and 3, prove the result holds for n=k+1, with the assumption that it's true for n=k
4. steps 2 and 3 tells you that
if it's true for n=1(from step 1), then it must also be true for n=1+1=2
so result true for n =2, then it must also be true for n=2+1=3
so result true for n =3, then it must also be true for n=3+1=4
...
Therefore the result holds true for all positive integers(in this case)

 

[funnytomato]

Thanks guys - eh trebla how did you get the 2nd line of the LHS side? I also don't get the (k+1)th term

The (k+1)(k+2)... =2^k... (2nd line) is the 2nd step, which makes the assumption of result being true for n=k. So just substitue k for n and get this equation

And the next step(step 3) is just some algebra to restore/find an expression similar to the LHS of assumption so that we can substitute RHS for it (I think this is the only step that requires you to think) , and then simplify to make it more apparent that it's ture for n=k+1

 

[Trebla]

Thanks guys - eh trebla how did you get the 2nd line of the LHS side? I also don't get the (k+1)th term

2k + 2 = 2(k + 1)
and then I moved it to the left of the expression...

This is the statement for n = k



This is the statement for n = k + 1


which simplifies to

 

[Drongoski]

Nicely explained.