# STEP 3 paper helpWatch

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Thread starter 5 years ago
#1
Hi, I'm new so apologies if this doesn't come out too well.

The question is as follows:

Show that B= .

I've expanded out the binomial in B and simplified, but I'm stuck at the very end.

. It's this last step, and i have a feeling i will be kicking myself when I see it or get a nudge in the right direction. Edit: The are fixed numbers.
1
5 years ago
#2
(Original post by maths learner)
Hi, I'm new so apologies if this doesn't come out too well.

The question is as follows:

Show that B= .

I've expanded out the binomial in B and simplified, but I'm stuck at the very end.

. It's this last step, and i have a feeling i will be kicking myself when I see it or get a nudge in the right direction.
You just need to substitute back for the definition of A in the middle term on your RHS (i.e. write the sum of x_n in terms of A).
0
Thread starter 5 years ago
#3
(Original post by davros)
You just need to substitute back for the definition of A in the middle term on your RHS (i.e. write the sum of x_n in terms of A).
I feel dumb. I can't get it to cancel...
0
5 years ago
#4
(Original post by maths learner)
I feel dumb. I can't get it to cancel...
You have so ...
0
Thread starter 5 years ago
#5
(Original post by davros)
You have so ...
Where did the come from? Sorry it's been a long day (I shouldn't try to do STEP when half asleep).
0
5 years ago
#6
Note: it's more obvious if you write the middle term as
1
5 years ago
#7
Note also that this is basically the proof that if you've covered that in stats.
0
5 years ago
#8
(Original post by maths learner)
Where did the come from? Sorry it's been a long day (I shouldn't try to do STEP when half asleep).
It's (-2A/n)(nA).

(I was trying to hint at it subtly, but perhaps too subtly! Anyway, I'm off to watch the Dr Who thing now!)
1
Thread starter 5 years ago
#9
(Original post by DFranklin)
Note: it's more obvious if you write the middle term as
Ohhh yeah. Ugh I need to get use to writing my fractions in a form like that, I get tripped up like that quite often. Thanks! Also Yeah I have done the stats thing .
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Thread starter 5 years ago
#10
(Original post by DFranklin)
Note: it's more obvious if you write the middle term as
Hi, I'm still struggling away at question 1 on the STEP 2010 paper, I was wondering if you would mind helping again...

we have the following sums:

I need to express D in terms of B, A and n. And then show that for all values of but that D<B if and only if .

I'm on the expressing D in terms of the other sums part I've got the following so far:

(By the same logic as how B was expressed above).
. Which was found by rearranging the B term and rewriting C in terms of A. Now i'm stuck...
0
5 years ago
#11
I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
0
Thread starter 5 years ago
#12
(Original post by DFranklin)
I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
I'll have a go now. Thanks for the reply. I don't see from first glance how we just get nB though. Opps missed you wrote and other terms XD!
0
5 years ago
#13
(Original post by maths learner)
Hi, I'm still struggling away at question 1 on the STEP 2010 paper, I was wondering if you would mind helping again...

we have the following sums:

I need to express D in terms of B, A and n. And then show that for all values of but that D<B if and only if .

I'm on the expressing D in terms of the other sums part I've got the following so far:

(By the same logic as how B was expressed above).
. Which was found by rearranging the B term and rewriting C in terms of A. Now i'm stuck...
I almost agree with you but I have
for the first two terms, otherwise I agree with you.
0
Thread starter 5 years ago
#14
(Original post by DFranklin)
I haven't checked your algebra, but the obvious next step is to multiply by (n+1), which will give you (n+1)D on the LHS and nB on the RHS (plus some other terms).

So at that point you'll need to show the sum of those other terms is non-negative. Again, I haven't done the algebra, but it should be straightforward.
Okay so I've simplified down, and gotten to:

. Which is what is gotten in the STEP solution guide. But then I don't know how you actually prove that for all values of .

I think I'm getting there. For the first part of the inequality can I simply say that for all values since it's obvious by looking at it... I'm not really sure how you prove it. For the second part I have this:

. Now this is less than B iff . But then I'm not too sure. I know if I expanded out rather than re wrote the term containing A in it I can start to get some form of quadratic which I presume I'll need, so that could be an approach?
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Thread starter 5 years ago
#15
(Original post by brianeverit)
I almost agree with you but I have
for the first two terms, otherwise I agree with you.
hmm. Maybe I made a slip up or something, but anyway I end up with what they get in the solution, I just don't know how you prove the inequalities :/.
0
5 years ago
#16
(Original post by maths learner)
hmm. Maybe I made a slip up or something, but anyway I end up with what they get in the solution, I just don't know how you prove the inequalities :/.
See attatched
0
Thread starter 5 years ago
#17
(Original post by brianeverit)
See attatched
Ah brilliant thank you. I'll work through it and see if I can follow the logic. I hate proving bi-conditional statements. Find them tricky.
0
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