STEP 2013 Solutions

Is it a valid to argue that since $\displaystyle u_{2n} = 1+1/(u_{2n-1})$
then if $\ u_{2n-1}$ tends to a limit then surely is $\ u_{2n}$
?

I think this is OK for Step 1 Q6

Don't quite understand S1,Q3, for why the part2's answer is not "and" but "or?

What it means is (X*(Y*Z)) is only ever equal in value to ((X*Y)*Z for ANY points X,Y,Z when lambda takes specific values...find those values!!

Let me explain...Work out X*(Y*Z) for x,y,z using the given expression for X*Y
Similarly for (X*Y)*Z for x,y,z
Put the two expressions you get equal then what restrictions does this place on lambda?? (so it will work for ANY x,y,z)
When you have the values it means that the only time the two original expressions in X,Y,Z are equal for all values is
the two values you have found.
Any other values of lambda means that the two expressions in X,Y,Z will be distinct for all values of x,y,z except the trivial
case x=y=z=0.

but , for example, lambda=0.1 and x=z, it satisfy "lamda<>0, 1 or X <> Z", but they are not distinct, so I think it must satisfy "lambda<>0, 1 and X <>Z" ?

exactly right....I was hoping you would see this extra condition...thats what gets the extra marks in STEP!!