Cambridge HSC MX1 Textbook Marathon/Q&A (6 Viewers)

123ryoma12 · Sep 30, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand · Sep 30, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

123ryoma12 said:
View attachment 33487

$\noindent Recall that the binomial coefficients are symmetric about the central binomial coefficient $\binom{2n}{n}$. Hence if we take the sum of coefficients from $r=0$ to $n-1$, it's half what we'd get if we took the sum of \textsl{all} $\binom{2n}{r}$ from $r = 0$ to $2n$ excluding the middle term $r=n$. In other words, $ \sum _{r=0}^{n-1} \binom{2n}{r} = \frac{1}{2}\left( \sum _{r=0}^{2n} \binom{2n}{r} - \binom{2n}{n}\right)$. Now, $\sum _{r=0}^{2n}\binom{2n}{r} = 2^{2n}$ (by subbing $x=1$ into the binomial expansion), so we have $\sum _{r=0}^{n-1}\binom{2n}{r} = \frac{1}{2}\left(2^{2n} - \binom{2n}{n}\right)$. Thus$

$\begin{align*}\sum _{r=0}^{n}\binom{2n}{r} &= \frac{1}{2}\left(2^{2n} - \binom{2n}{n}\right) + \binom{2n}{n} \\ &= \frac{1}{2}\left(2^{2n} + \binom{2n}{n}\right)\\ &= 2^{2n-1} + \frac{(2n)!}{2(n!)^{2}}.\end{align*}$

unknownteenager · Nov 29, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

Hi, is anyone able to do question 17 d from exercise 10E ?
Thank you!

Lugia101101 · Dec 1, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

If N is to the left of U, the answers must all be in the form ____N____U____, where the underscores represent places the remaining four letters can go. So:
Consider _N_U_, where we are trying to insert M. There are 3 places for it to go.
Now consider what will be left, we have: _N_U_M_, or _N_M_U_, or _M_N_U_. Regradless of where we place the M, there are four places for the next letter.
Following this, wherever we place the next letter will take away one available space, but create two either side of it, netting one more. So for M there are 3, for B there are 4, for E there are 5, and for R there are 6.
Doing it this way gives an answer of 3*4*5*6=360, and with this method we cover the event of placing the letters down in different orders.

aycaramba · Dec 6, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

http://prnt.sc/dfvluh pls halp. what does this ques mean

Lugia101101 · Dec 6, 2016

Re: Year 12 Mathematics 3 Unit Cambridge Question & Answer Thread

It's asking what values of a, b, and c are required to change the degree of the polynomial. For example, if you had a polynomial $ax^3+bx^2+cx+d$ , and you wanted a polynomial of degree 1, a and b would both have to be zero, so that the polynomial became $cx+d$ , with the degree of a polynomial being the highest exponent.

sleepywood · Dec 25, 2017

Can anyone help me question 23 of Exercise 1 E

Drongoski · Dec 25, 2017

sleepywood said:
Can anyone help me question 23 of Exercise 1 E

Yr 11 or Yr 12 text? Edition? Is it the one involving y = sin^-1 x ??

Cambridge HSC MX1 Textbook Marathon/Q&A (6 Viewers)

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