STEP 3 paper help

Hi, I'm new so apologies if this doesn't come out too well.

The question is as follows:
$A=\frac{1}{n}\displaystyle\sum_{k=1}^n x_k$
$B=\frac{1}{n}\displaystyle\sum_{k=1}^n (x_k-A)^2$

Show that B= $\frac{1}{n}\displaystyle\sum_{k=1}^n (x_k^2 -A^2)$.

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

$B=\frac{1}{n}\displaystyle\sum_{k=1}^n (x_k-A)^2 \Rightarrow \frac{1}{n}\displaystyle\sum_{k=1}^n x_k^2 - \frac{2A}{n}\displaystyle\sum_{k=1}^n x_k + A^2$. 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).

I feel dumb. I can't get it to cancel...

You have $-2A^2 + A^2$ so ...

Where did the $-2A^2$ come from? Sorry it's been a long day (I shouldn't try to do STEP when half asleep).

Note: it's more obvious if you write the middle term as $\displaystyle 2A \frac{1}{n} \sum_{k=1}^n x_k$

Note: it's more obvious if you write the middle term as $\displaystyle 2A \frac{1}{n} \sum_{k=1}^n x_k$

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.

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:

$A=\frac{1}{n}\sum\limits_{k=1}^nk_x$
$B=\frac{1}{n}\sum\limits_{k=1}^n (x_k-A)^2$
$C=\frac{1}{n+1}\sum\limits_{k=1}^{n+1} x_k$
$D=\frac{1}{n+1}\sum\limits_{k=1}^{n+1} (x_k -C)^2$

I need to express D in terms of B, A $x_{n+1}$ and n. And then show that $(n+1)D\geq nB$ for all values of $x_{n+1}$ but that D<B if and only if $A- \sqrt\frac{(n+1)B}{n} < x_{n+1} < A + \sqrt\frac{(n+1)B}{n}$.

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

$D=\frac{1}{n+1}\sum\limits_{k=1}^{n+1} x_k^2 -C^2$ (By the same logic as how B was expressed above).
$D=\frac{1}{n+1}(n(B+A^2)+x_{n+1}^2) - \frac{(nA+x_{n+1})^2}{(n+1)^2}$. Which was found by rearranging the B term and rewriting C in terms of A. Now i'm stuck...

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 almost agree with you but I have
$D=\frac{1}{n+1}B+\frac{n}{(n+1)^2}A^2$ 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 :/.

See attatched