Jump to content
Sign in to follow this  

Math HL (Mathematical induction) Can someone kindly explain this question to me.

Recommended Posts

I am not sure what exactly you are confused about so please follow up.

There are three steps to induction. 1) show for first case 2) Assume true for some n=k. 3) show that it is true for all n=k+1. The idea is that if we are successful, we show n=1 and that implies n=2, n=3 ... are all valid. 

We know that k is an element of Z^+, so first case is k = 1(there have been questions where they did k is element of N, which would start with k=0, just be careful). 

The key is to substitute your assumption into your third step. Easiest way is to expand (k+1)^3 then substitute. Then they showed that k^2+k+2 is even, which would make 3(k^2+k+2) is a multiple of 6.

Share this post

Link to post
Share on other sites

Honestly, there are 5 steps to Mathematical induction:

1. show that p(n) is true for any value of n

2. assume p(k) is true

3. consider p(k+1) 

4. show that p(k+1) is true

5. conclusion

for this question, you are trying to prove that p(n) is always a multiple of 6 for n is an integer; so for any value of n, p(n) can be written as 6 times any random integer, which is exactly what is meant when something is a multiple of 6. The rest of the steps are pretty self-explanatory. 

Share this post

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Sign in to follow this  

  • Create New...