STEP II 2014 8

From the binomial expansion: c_r=\binom{n}{r}a^{n-r}b^r=\frac{n!}{r!(n-r)!}a^{n-r}b^r
Assuming m\not=0, c_m\geq c_{m-1}:

\begin{align} c_m&\geq c_{m-1}\\ \frac{n!}{m!(n-m)!}a^{n-m}b^m&\geq\frac{n!}{(m-1)!(n-m+1)!}a^{n-m+1}b^{m-1}\\ b(n-m+1)&\geq am\\ b(n+1)&\geq am+bm\\ \frac{b(n+1)}{a+b}&\geq m\\ \end{align}

Since n,a,b>0, \frac{b(n+1)}{a+b}>0, so the result also holds for m=0.

Similarly, assuming m\not=n:

\begin{align} c_m&\geq c_{m+1}\\ \frac{n!}{m!(n-m)!}a^{n-m}b^m&\geq\frac{n!}{(m+1)!(n-m-1)!}a^{n-m-1}b^{m+1}\\ a(m+1)&\geq b(n-m)\\ am+bm&\geq bn-a\\ m&\geq \frac{bn-a}{a+b}\\ m&\geq \frac{bn-a}{a+b}+1-1\\ m&\geq \frac{bn-a}{a+b}+\frac{a+b}{a+b}-1\\ m&\geq \frac{b(n+1)}{a+b}-1\\ \end{align}

For the special case n=m, we first note that \frac{b}{a+b}<1:

\begin{align} 1&\geq \frac{b}{a+b}\\ n+1&\geq \frac{b(n+1)}{a+b}\\ n&\geq \frac{b(n+1)}{a+b}-1\\ m&\geq \frac{b(n+1)}{a+b}-1\\ \end{align}

so the result still holds.
Combining these two inequalities gives the stem result.

Since k-1\leq m\leq k for some k, m can take at most 2 integral values.

For the next parts, we start by noting that G(n,a,b)=\left\lfloor\frac{b(n+1)}{a+b}\right\rfloor.
part (i)

\begin{align} G(9,1,3)&=\left\lfloor\frac{3(9+1)}{1+3}\right\rfloor=7\\ G(9,2,3)&=\left\lfloor\frac{3(9+1)}{2+3}\right\rfloor=6 \end{align}

part (ii)

\begin{align} G(2k,a,a)&=\left\lfloor\frac{a(2k+1)}{a+a}\right\rfloor\\ &=\left\lfloor\frac{2k+1}2\right\rfloor\\ &=k\\\\ G(2k-1,a,a)&=\left\lfloor\frac{a(2k-1+1)}{a+a}\right\rfloor\\ &=\left\lfloor\frac{2k}2\right\rfloor\\ &=k \end{align}

part (iii)
In order to maximise G(n,a,b)=\left\lfloor\frac{b(n+1)}{a+b}\right\rfloor, we want to minimise the denominator. This is achieved when a=1, since a is a positive integer.
part (iv)
Using the fact that \frac{b}{1+b}<1:

\begin{align} G(n,1,b)&=\left\lfloor\frac{b(n+1)}{1+b}\right\rfloor\\ &=\left\lfloor\frac{b}{1+b}(n+1)\right\rfloor\\ &<n+1 \end{align}

Therefore, G(n,1,b) is at most n. This is achieved when:

\begin{align} \left\lfloor\frac{b(n+1)}{1+b}\right\rfloor&=n\\ \frac{b(n+1)}{1+b}&\geq n\\ b(n+1)&\geq n(1+b)\\ b&\geq n \end{align}
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