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

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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.

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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. 

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