2016 HSC Past Paper Question (1 Viewer)
- Thread starter Andy005
- Start date
Given that
 - kD(k-1) = -[D(k-1) - (k-1)D(k-2)] )
for some 
.
Set
to get
:\, 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
to get
:\, \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:
:\, D(n) - nD(n-1) = -[-[\color{blue}D(n-2) - (n-2)D(n-3)\color{black}]] )
Set
to get
:\, \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:
:\, D(n) - nD(n-1) = -[-[-[D(n-3) - (n-3)D(n-4)]]] )
If you look at (1), (3) and (5):
:\, D(n) - nD(n-1) = -[{D(n-1) - (n-1)D(n-2)}] )
:\, D(n) - nD(n-1) = -[-[D(n-2) - (n-2)D(n-3)]] )
It's pretty clear what is happening on the RHS.
If you keep repeating this process, then for any finite
you will eventually get
 - 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:
:\, D(n) - nD(n-1) = \color{red}-\color{black}[D(n-\color{red}1\color{black})-...] )
:\, D(n) - nD(n-1)= \color{red}-\color{black}[\color{red}-\color{black}[D(n-\color{red}2\color{black})-...] )
:\,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(n-(n-2)))
so when we eventually reduce that term on the RHS to  )
then there will be 
minus signs on the RHS.
That is,
 - 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 )
and  )
is not defined. Therefore we need 
.
Set
That red expression on the RHS looks very similar to the LHS.
Set
Substitute (2) back into (1) to get:
Set
Substitute (4) back into (3) to get:
If you look at (1), (3) and (5):
It's pretty clear what is happening on the RHS.
If you keep repeating this process, then for any finite
So how many minus signs are on the RHS? Well let's look at those equations again:
That red number happens to be the same as the number of minus signs on the RHS.
Now
That is,
Note that the LHS contains
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