The Student Room Group

Proof by induction

I keep getting stuck on inductive step! Any tips ?!?
Reply 1
image.jpgHow did he get from line 1 to 2 on the inductive step ?! What did he factor out
Original post by MM2002
How did he get from line 1 to 2 on the inductive step ?! What did he factor out


Leave the -1 alone and factorise the leftover two terms by taking out (k+1)!
Reply 3
Original post by RDKGames
Leave the -1 alone and factorise the leftover two terms by taking out (k+1)!

Am I left with this expression then (k+1)! [ -1 +(k+1)]
Reply 4
I did that also. I still can not get the k+2 part
Original post by MM2002
Am I left with this expression then (k+1)! [ -1 +(k+1)]


No.

(k+1)! + (k+1)(k+1)! = (k+1)! [ 1 + (k+1)]
Reply 6
Original post by RDKGames
No.

(k+1)! + (k+1)(k+1)! = (k+1)! [ 1 + (k+1)]

I don't see how the one is positive?
Original post by MM2002
I don't see how the one is positive?


Expand the brackets then to verify that it must be +1 and not -1.

If its -1 then expanding back will give you -(k+1)! which is not a term we have.
Reply 8
Original post by RDKGames
Expand the brackets then to verify that it must be +1 and not -1.

If its -1 then expanding back will give you -(k+1)! which is not a term we have.

But, by factoring I don't see why it turns to -1
Reply 9
Original post by MM2002
image.jpgHow did he get from line 1 to 2 on the inductive step ?! What did he factor out

(k+1)! - 1 + (k+1)(k+1)!
= [(k+1)! + (k+1)(k+1)!] - 1
=[ { (k+1)! } { (1) + (k+1) } ] - 1
= [(k+1)! (k+2)] - 1
= (k+2)! - 1

does that help?
Original post by MM2002
But, by factoring I don't see why it turns to -1


It doesnt ??

You seem very confused.

I have just shown you the factorisation, and then you asked why its +1 and not -1, and now you ask how it turns into -1... which it doesnt.


If you are talking about the -1 at the end of the expression on its own, then this is the same -1 I told you to ignore in my first post. We dont manipulate it at all here.
(edited 4 years ago)
Reply 11
Original post by shreytib
(k+1)! - 1 + (k+1)(k+1)!
= [(k+1)! + (k+1)(k+1)!] - 1
=[ { (k+1)! } { (1) + (k+1) } ] - 1
= [(k+1)! (k+2)] - 1
= (k+2)! - 1

does that help?


AMAZING!! Finally, understand how all the values come about. Thanks sooo much! Also, any tips on solving the inductive step?
Reply 12
Original post by RDKGames
It doesnt ??

You seem very confused.

I have just shown you the factorisation, and then you asked why its +1 and not -1, and now you ask how it turns into -1... which it doesnt.


If you are talking about the -1 at the end of the expression on its own, then this is the same -1 I told you to ignore in my first post. We dont manipulate it at all here.

That clears things up. I was thinking about the -1 you told me ignore.

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