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I'm goin for 4u and 3u

how did u get 6! since it has only 3 spots. You need to take case or atleast i did. case 1 when only 1b and 2c are on block 1 case 2 when there is 2B and 1C then when either all 3B or 3C are on first block

Yes but many students will probably be familiar with it without even knowing that its called a telescoping sum. i learnt it when i saw it quoted on the forums so i researched it.

my answer was 27 and 56. I thought of it as 3 seperate blocks, first block has 2B and 1C second has 1A and 2C then third has 2A and 1B, this is the only arrangement that we obey the restriction and total arrangements is \dfrac{3!3!3!}{2!2!2!}=27 for part ii, if we take the case there was 1B...

making it in terms of x^2 is not neccesary but neater since you can substitute the equation without having to square root which looks a bit messy with the plus or minus sign, you end up squaring anyway though

Oh thats one i dont remember perfectly but when you differentiate, i remember getting \alpha^2=\dfrac{3q}{p} or something similar. From original equation you wanna make it interms of x^2 for example x(px^2-q)=-r \rightarrow x^2(px^2-q)^2=r^2 Subbing \alpha^2=\dfrac{3q}{p} gives 4p^3=qr^2 i...

Yes

For the telescoping sum, what happens is that the "middle terms" cancel out: \sum_{n=0}^{N}(a_n-a_{n+1})=(a_0-a_1)+(a_1-a_2)+(a_2-a_3)+...+(a_{n}-a_{n+1})=a_0-a_{n+1} As for part 3, because the magnitude of x is less than 1, the larger we "power it" the smaller it becomes. If we raise it to...

you mean the whole exam? i dont have it, safety period... I remember most of the exam though

Solution: $i)\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x^{2^n} (1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}=\dfrac{(1-x^{2^{n+1}})-(1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})} =\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}} $ii) Hence show that$ \sum_{n=0}^{N}...

pm'd

Il finish the sols here first then il move it lol

Solution: $i)\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x^{2^n} (1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}=\dfrac{(1-x^{2^{n+1}})-(1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})} =\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}} $ii) Hence show that$ \sum_{n=0}^{N}...

That polynomial with the roots of unity i have posted on the 4u marathon a while ago

Lol i got a good memory when it comes to math

Yeh it was exactly that, a telescoping sum

$i) Show that$ \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}} $ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}} $iii)Given that$ \lim_{n\rightarrow \infty}\sum_{n=0}^{N}...

yes il post it from memory

You use the previous part by substituting x=2014^{-1}

\; $Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}} Which i believe is \dfrac{1}{2013}