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C2H6O

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Could you post your answers here if you have time sir? Thanks!
 

Sam14113

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Nice to see you doing this again :)
Your concern about 16 (b) (ii) caught my attention -- I haven't thought too much about it and I may have made a mistake but I think it's really just a case of induction.

If you consider the sequence of unordered sets , you can show by induction that for all integers . The base case is clear and the induction step is effectively the exact logic you showed.

In my opinion it's quite unclear how much of that would be required for what can surely only be 2 marks. I suppose we'll find out.
 

epicmaths

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Nice to see you doing this again :)
Your concern about 16 (b) (ii) caught my attention -- I haven't thought too much about it and I may have made a mistake but I think it's really just a case of induction.

If you consider the sequence of unordered sets , you can show by induction that for all integers . The base case is clear and the induction step is effectively the exact logic you showed.

In my opinion it's quite unclear how much of that would be required for what can surely only be 2 marks. I suppose we'll find out.
Yes induction like you said works out quite quickly...I tried to be special and uploaded a solution involving modulo arithmetic, conceding that induction is probably what they wanted.
 

Sam14113

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Yes induction like you said works out quite quickly...I tried to be special and uploaded a solution involving modulo arithmetic, conceding that induction is probably what they wanted.
Fair
 

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