# Thread: 2016 HSC Past Paper Question

1. ## 2016 HSC Past Paper Question

Screen Shot 2018-09-07 at 7.40.09 pm.png

Q(IV) I don't exactly understand the solution to this question.

Thanks.

2. ## Re: 2016 HSC Past Paper Question

Given that

$D(k) - kD(k-1) = -[D(k-1) - (k-1)D(k-2)]$ for some $k > 2$.

Set $k =n$ to get

$(1):\, D(n) - nD(n-1) = -[\color{red}{D(n-1) - (n-1)D(n-2)}\color{black}]$

That red expression on the RHS looks very similar to the LHS.

Set $k = n-1$ to get

$(2):\, \color{red}{D(n-1) - (n-1)D(n-2)} \color{black}= -[D(n-2) - (n-2)D(n-3)]$

Substitute (2) back into (1) to get:

$(3):\, D(n) - nD(n-1) = -[-[\color{blue}D(n-2) - (n-2)D(n-3)\color{black}]]$

Set $k = n-2$ to get

$(4):\, \color{blue}{D(n-2) - (n-2)D(n-3)} \color{black}= -[D(n-3) - (n-3)D(n-4)]$

Substitute (4) back into (3) to get:

$(5):\, D(n) - nD(n-1) = -[-[-[D(n-3) - (n-3)D(n-4)]]]$

If you look at (1), (3) and (5):

$(1):\, D(n) - nD(n-1) = -[{D(n-1) - (n-1)D(n-2)}]$
$(3):\, D(n) - nD(n-1) = -[-[D(n-2) - (n-2)D(n-3)]]$
$(5):\, D(n) - nD(n-1) = -[-[-[D(n-3) - (n-3)D(n-4)]]]$

It's pretty clear what is happening on the RHS.

If you keep repeating this process, then for any finite $n$ you will eventually get

$D(n) - nD(n-1) = -[-[-[...-[D(2) - (2)D(1)]]]] = -[-[-[...-[1]]]]$

So how many minus signs are on the RHS? Well let's look at those equations again:

$(1):\, D(n) - nD(n-1) = \color{red}-\color{black}[D(n-\color{red}1\color{black})-...]$
$(3):\, D(n) - nD(n-1)= \color{red}-\color{black}[\color{red}-\color{black}[D(n-\color{red}2\color{black})-...]$
$(5):\,D(n) - nD(n-1)= \color{red}-\color{black}[\color{red}-\color{black}[\color{red}-\color{black}[D(n-\color{red}3\color{black})-...]$

That red number happens to be the same as the number of minus signs on the RHS.

Now $D(2) = D(n-(n-2))$ so when we eventually reduce that term on the RHS to $D(2)$ then there will be $n-2$ minus signs on the RHS.

That is,

$D(n) - nD(n-1) = \underbrace{-[-[-[}_{n-2\,\,\mathrm{times}} ... [1] ]]] = (-1)^{n-2} = (-1)^n \cdot \frac{1}{(-1)^{2}} = (-1)^n$

Note that the LHS contains $D(n-1)$ and $D(0)$ is not defined. Therefore we need $n -1 > 0 \implies n > 1$.

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