Problems

Filters
Clear Filters

2 problems found

2014 Paper 2 Q8
D: 1600.0 B: 1486.3
binomial theorem binomial coefficients optimisation inequality integer constraints maximisation positive integer n

For positive integers \(n\), \(a\) and \(b\), the integer \(c_r\) (\(0\le r\le n\)) is defined to be the coefficient of \(x^r\) in the expansion in powers of \(x\) of \((a+bx)^n\). Write down an expression for \(c_r\) in terms of \(r\), \(n\), \(a\) and \(b\). For given \(n\), \(a\) and \(b\), let \(m\) denote a value of \(r\) for which \(c_r\) is greatest (that is, \(c_m \ge c_r\) for \(0\le r\le n\)). Show that \[ \frac{b(n+1)}{a+b} - 1 \le m \le \frac {b(n+1)}{a+b} \,. \] Deduce that \(m\) is either a unique integer or one of two consecutive integers. Let \(G(n,a,b)\) denote the unique value of \(m\) (if there is one) or the larger of the two possible values of \(m\).

  1. Evaluate \(G(9,1,3)\) and \(G(9,2,3)\).
  2. For any positive integer \(k\), find \(G(2k,a,a)\) and \(G(2k-1,a,a)\) in terms of \(k\).
  3. For fixed \(n\) and \(b\), determine a value of \(a\) for which \(G(n,a,b)\) is greatest.
  4. For fixed \(n\), find the greatest possible value of \(G(n,1,b)\). For which values of \(b\) is this greatest value achieved?

Solution: \(c_r = \binom{n}{r}a^{n-r}b^r\) \begin{align*} && c_m &\geq c_{m+1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m+1} a^{n-m-1}b^{m+1} \\ \Rightarrow && \frac{1}{(n-m)}a &\geq \frac{1}{m+1}b \\ \Rightarrow && (m+1)a &\geq (n-m)b \\ \Rightarrow && m(a+b) &\geq nb -a \\ \Rightarrow && m &\geq \frac{n(b+1)-a-b}{a+b}=\frac{n(b+1)}{a+b} - 1 \\ \\ && c_m &\geq c_{m-1} \\ \Rightarrow && \binom{n}{m} a^{n-m}b^m &\geq \binom{n}{m-1} a^{n-m+1}b^{m-1} \\ \Rightarrow && \frac{1}m b &\geq \frac{a}{(n-m+1)} \\ \Rightarrow && (n-m+1)b &\geq ma \\ \Rightarrow && (n+1)b &\geq m(a+b) \\ \Rightarrow && m &\leq \frac{(n+1)b}{a+b} \end{align*} Since \(m\) lies between two values \(1\) apart is is either equal to one of those two values or is the unique integer between them. Let \(\displaystyle G(n,a,b) = \left \lfloor \frac{b(n+1)}{a+b} \right \rfloor\), so

  1. \(\,\) \begin{align*} && G(9,1,3) &= \left \lfloor \frac{3(9+1)}{1+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{4} \right \rfloor \\ &&&= 7 \\ \\ && G(9,2,3) &= \left \lfloor \frac{3(9+1)}{2+3} \right \rfloor \\ &&&= \left \lfloor \frac{30}{5} \right \rfloor \\ &&&= 6 \end{align*}
  2. \(\,\) \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 k\right \rfloor \\ &&&= k \\ \end{align*}
  3. \(G(n,a,b)\) is decreasing in \(a\), therefore take \(a = 1\).
  4. For fixed \(n\), we are looking at \(\frac{a(n+1)}{a+b}\) and we want this to be as large as possible. By considering \((n+1) - \frac{b(n+1)}{a+b}\) we can see this is increasing in \(b\) and the largest value possible is \(n\). This is achieved when \begin{align*} && \frac{b(n+1)}{a+b} & \geq n \\ \Leftrightarrow && bn + b &\geq na + bn \\ \Leftrightarrow && b& \geq na \end{align*}

View
1989 Paper 1 Q5
D: 1500.0 B: 1516.0
binomial theorem binomial coefficients substitution summation even and odd terms differentiation positive integer n

Write down the binomial expansion of \((1+x)^{n}\), where \(n\) is a positive integer.

  1. By substituting particular values of \(x\) in the above expression, or otherwise, show that, if \(n\) is an even positive integer, \[ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{n}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots+\binom{n}{n-1}=2^{n-1}. \]
  2. Show that, if \(n\) is any positive integer, then \[ \binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n2^{n-1}. \]
Hence evaluate \[ \sum_{r=0}^{n}\left(r+(-1)^{r}\right)\binom{n}{r}\,. \]

Solution: \[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

  1. \begin{align*} (1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\ (1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\ &= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} \end{align*} Therefore \(\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} = \frac{2^n}{2} = 2^{n-1}\)
  2. \begin{align*} && (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\ x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k} \end{align*} as required
\begin{align*} \sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1} \end{align*}

View