STEP II - Q8 (LaTeX)
{n,a,b,cr}Z+cr,(0≤r≤n)=coefficient of xr in (a+bx)ncr=nCr⋅an−rbr=r!(n−r)!n!⋅an−rbrcm≥cr for 0≤r≤nIf
cm is a maximum, it must be
≥cm+1 and
≥cm−1Taking first,
cm+1 :
(m+1)!(n−(m+1))!n!an−(m+1)b(m+1)≤m!(n−m)!n!an−mbmDivide both sides by the common factor
m!(n−(m+1))!n!an−(m+1)bmGiving
m+1b≤n−mabn−bm≤am+abn−a≤am+bma+bbn−a≤mThis is almost the LHS of what we are required to show, just needs a little fiddling
a+bbn+b−(a+b)≤ma+bb(n+1)−1≤m(∗∗)Now repeat the process for
cm−1:
(m−1)!(n−(m−1))!n!an−(m−1)b(m−1)≤m!(n−m)!n!an−mbmDivide both sides by the common factor
(m−1)!(n−m)!n!an−mbm−1Giving
n−(m−1)a≤mbam≤bn+b−bmm(a+b)≤b(n+1)m≤a+bb(n+1)Which combines with (**) to give the required solution
a+bb(n+1)−1≤m≤a+bb(n+1)Considering the possible values of
mLet y=a+bb(n+1)y−1≤m≤yEither
y and
y−1 are both integers, or
y and
y−1 are non-integer values either side of an integer, so if
m is an integer, it is either
y and
y−1 (2 values) or it lies between them (1 value).
For the next parts, we are asked to find either the value of
m where there is 1 value, or the larger value if there are 2. This means we can consider the function
G(n,a,b) to produce the
Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is
⌊.
G(n,a,b)=⌊a+bb(n+1)i)
Unparseable latex formula:G(9,1,3) = \lfloor{\frac{3(9+1)}{1+3}} = \lfloor\frac{30}{4} = \floor{7.5} = 7
G(9,2,3)=⌊2+33(9+1)=⌊530=6ii)
G(2k,a,a)=⌊a+aa(2k+1)=⌊22k+1=⌊(k+0.5)=kG(2k−1,a,a)=⌊a+aa(2k−1+1)=⌊22k=kiii)
If we fix
n and
b, the only part of
G that can alter is the denominator containing
a, so for large
G we want small
a, so
Gmax is when
a=1. (Zero is not a positive integer).
iv)
For fixed
a=1 and (unknown) fixed
n, the part of the function varying is:
b+1b(∗∗∗)This tends to 1 as b tends to infinity, but will never actually reach 1, so
(∗∗∗)(n+1) tends towards but never reaches
n+1 so for fixed
nG(n,1,b)max=n when b=∞---
This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.