### Absurdity Again

What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?

### Ball Bearings

If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.

### Overarch 2

Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?

# Generally Geometric

##### Age 16 to 18 Challenge Level:

Andy from Clitheroe Royal Grammar School sent us his work on this problem. He's given us two methods; can you see why he prefers the second one?

We begin by summing the series

$x+2x^2+3x^3+4x^4+\cdots$

 $x$ + $x^2$ + $x^3$ + $x^4$ + $\cdots$ + $x^2$ + $x^3$ + $x^4$ + $\cdots$ + $x^3$ + $x^4$ + $\cdots$ + $\cdots$

In other words, we are writing it as a sum of geometric series!

Now, let us factorise the above sum as follows:

$(x + x^2 + x^3 + x^4+\ldots)(1 + x + x^2 + x^3 + x^4+\ldots)$

Wow, a product of geometric series!

We can then take a factor of $x$ out the first bracket to leave us with

$x(1 + x + x^2 + x^3+\ldots)^2$

Using the geometric sum given in the question, this comes to $$x\times \left(\frac{1}{1-x}\right)^2 = \frac{x}{(1-x)^2}$$ __

A similar method could be used for the series $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, factorising it as $(x + 3x^2 + 5x^3 + 7x^4+\ldots)(1 + x + x^2 + x^3 +\ldots)$, then writing the left hand bracket as $(x + x^2 + x^3+\ldots + 2x^2 + 4x^3 + 6x^4+\ldots)$, from which point we can use our previous sum to obtain an answer. Unfortunately this doesn't generalise easily into higher powers, the amount of working needed growing much larger at each stage.

A more elegant solution is differentiation. If we differentiate our first series, we get $1 + 4x + 9x^2 + 16x^3+\ldots + n^2x^{n-1}+\ldots$. Multiplying through by $x$ gives us $x + 4x^2 + 9x^3+\ldots + n^2 x^n+\ldots$, which is the $n^2 x^n$ series we need.

If $x + 2x^2 + 3x^3 + 4x^4+\ldots = x/(1-x)^2$ then $x(d[x + 2x^2 + 3x^3+\ldots]/dx) = x(d[x/(1-x)^2]/dx)$.

But the left-hand side is equal to $x + 4x^2 + 9x^3 + 16x^4 +\ldots$, the sequence we want to sum.

We can resolve the right-hand using the quotient rule, and it comes to $x(1+x)/(1-x)^3$.

__

To take it into higher powers, notice that

$d[x + 4x^2 + 9x^3+\ldots]/dx = 1 + 8x + 27x^2+\ldots$.

Therefore $x d[x + 4x^2 + 9x^3+\ldots]/dx = x + 8x^2 + 27x^3+\ldots$, our next sequence. We can differentiate the previous infinite sum and multiply by $x$ at each stage to get the sum for the next power, and by applying the same process to the closed-form expression, we can obtain a closed-form expression for the next power.