Binomial and SHM (1 Viewer)

pikachu975 · Jul 16, 2017

http://prntscr.com/fw395w

How would you do iii? It's from our school trial (copy pasted from bored of studies trial 2012) but I don't get the solution (2 parts because it's too big):

http://prntscr.com/fw39ki
http://prntscr.com/fw39oo

Thanks guys

EDIT: Also this simple harmonic motion one - http://prntscr.com/fw3ffl

Thanks!

InteGrand · Jul 16, 2017

pikachu975 said:
http://prntscr.com/fw395w

How would you do iii? It's from our school trial (copy pasted from bored of studies trial 2012) but I don't get the solution (2 parts because it's too big):

http://prntscr.com/fw39ki
http://prntscr.com/fw39oo

Thanks guys

EDIT: Also this simple harmonic motion one - http://prntscr.com/fw3ffl

Thanks!

For the SHM one, the answer must be (D) — this is the only option with the correct units for a period (seconds).

Note from the equation v^2 = a^2 - b^2 x^2, the LHS has units m^2 s^(-2). So the RHS does do. Thus a must have units m/s. Also, b^2 x^2 must have units m^2 s^(-2), so bx has units m/s. Since x has units m, b has units s^(-1).

InteGrand · Jul 16, 2017

pikachu975 said:
http://prntscr.com/fw395w

How would you do iii? It's from our school trial (copy pasted from bored of studies trial 2012) but I don't get the solution (2 parts because it's too big):

http://prntscr.com/fw39ki
http://prntscr.com/fw39oo

Thanks guys

EDIT: Also this simple harmonic motion one - http://prntscr.com/fw3ffl

Thanks!

$\noindent For the binomial sum, recall the following general series fact:$

$$\sum_{r = 0}^{n}a_{r} = \sum_{r = 0}^{n} a_{n-r}$.$

$\noindent (To see why this is true, you can write out both sums in long notation and you'll see that one is just written in reverse order of the other, with exactly the same terms.) Thus, remembering that $\binom{n}{r} = \binom{n}{n-r}$, we have$

$\begin{align*}S:= \sum_{r = 0}^{n} r \binom{n}{r}^{2} &= \sum_{r = 0}^{n} (n-r) \binom{n}{n-r}^{2} \\ &= n\sum_{r = 0}^{n} \binom{n}{n-r}^{2} - \sum_{r = 0}^{n} r \binom{n}{n-r}^{2}\\ &= n \sum_{r = 0}^{n} \binom{n}{r}^{2} - \underbrace{\sum_{r = 0}^{n} r \binom{n}{r}^{2}}_{= S} \\ \Rightarrow S &= n\binom{2n}{n} - S \quad (\text{by (ii)}) \\ \Rightarrow S &= \frac{n}{2}\binom{2n}{n}. \end{align*}$

InteGrand · Jul 16, 2017

pikachu975 · Jul 16, 2017

Thanks Integrand helped heaps

Binomial and SHM (1 Viewer)

pikachu975

Premium Member

InteGrand

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InteGrand

Well-Known Member

InteGrand

Well-Known Member

pikachu975

Premium Member

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