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Binomial expansion [2]

Largest coefficient

B2, 2011

Show that the power of \(x\) with the largest coefficient in the polynomial \(\displaystyle (1 + 2x/3)^{20}\) is 8. In other words, if we write the given polynomial as \( \sum_i a_ix^i\) then the largest coefficient \(a_i\) is \(a_8\).


The coefficient \( a_i = \binom{20}{i} \left(\frac{2}{3}\right)^i \). Consider the ratio of two consecutive terms: \( a_{i+1}/a_i \). \begin{align} \text{ratio } &= \frac{2}{3} \times \left( \frac{20!}{20-i-1!i+1!}/\frac{20!}{20-i!i!} \right) \\ &\\ &= \frac{2}{3} \cdot \frac{20-i!i!}{20-i-1!i+1!} = \frac{2(20-i)}{3(i+1)} \end{align} The ratio \( a_{i+1}/a_i > 1\) up to \(i\leq 7\) and strictly less than 1 for \(i>7\). Hence, the sequence of coefficients is bitonic with the peak occurring at \(a_8\).

Coefficient ratio

A7, 2015

(i) By the binomial theorem \((\sqrt{2}+1)^{10}=\sum_{i=0}^{10} C_{i}(\sqrt{2})^{i},\) where \(C_{i}\) are appropriate constants. Write the value of \(i\) for which \(C_{i}(\sqrt{2})^{i}\) is the largest among the 11 terms in this sum.

(ii) For every natural number \(n,\) let \((\sqrt{2}+1)^{n}=p_{n}+\sqrt{2} q_{n},\) where \(p_{n}\) and \(q_{n}\) are integers. Calculate \(\lim _{n \rightarrow \infty}\left(\frac{p_{n}}{q_{n}}\right)^{10}\).


(i) \(i=6\). Consider the ratio:
This ratio is \(>1\) till \(i=5\) and \(<1\) from \(i=6\) onwards. Similar to problem B2, 2011.

(ii) 32. Using binomial expansion see that \((\sqrt{2}-1)^{n}=\pm\left(p_{n}-\sqrt{2} q_{n}\right),\) where the sign depends on the parity of \(n .\) As \(n \rightarrow \infty,(\sqrt{2}-1)^{n} \rightarrow 0\) since \((\sqrt{2}-1)<1 .\) Thus \(\left(p_{n}-\sqrt{2} q_{n}\right) \rightarrow 0\) and so \(\frac{p_{n}}{q_{n}} \rightarrow \sqrt{2}\)

Inequalities [4]

AM-GM inequality

A4, 2011

Given positive real numbers \( a_1, a_2, \ldots , a_{2011} \) whose product \(a_1 a_2 \cdots a_{2011}=1\), what can you say about their sum \( S = a_1 + a_2 + · · · + a_{2011} \)?

  • \( S\; \) can be any positive number.
  • \( 1 \leq S \leq 2011\).
  • \( 2011 \leq S \text{ and } S\;\) is unbounded above.
  • \( 2011 \leq S \text{ and } S\;\) is bounded above.

\( 2011 \leq S \text{ and } S\;\) is unbounded above.

The first inequality follows from AM-GM inequality. To see why \(S\) is unbounded, set \( a_1=n \), \(a_2=1/n\) and the rest of \( a_is\) to 1. The sum \(S>n\) for any \(n\).

AM-GM inequality II

B5, 2012

Suppose \(x+x^{-1}=\frac{\sqrt{5}+1}{2}\).

(a) For any real \(r,\) show that \(\left|r+r^{-1}\right| \geq 2 .\) What kind of a number is \(x\)?


Without loss of generality, we may assume that \(r > 0\). Now use the AM-GM inequality:
\[ \frac{ r+\frac{1}{r} }{2} \geq \sqrt{ r \cdot \frac{1}{r} } \] since \(x+x^{-1}=\frac{\sqrt{5}+1}{2}<2\), the number \(x\) must be a non-real complex number.

Combinatorial and calculus inequalities

B2b, 2021

Prove or disprove:
(i) \( 2^{40} > 20! \)
(ii) \( 1-1/x \leq \ln x \leq x-1 \) for all \(x>0\)

Solution (i) False.
\begin{align*} 20! &> 2\times3\times4\times5\times6\times 7\times 8\times8\times 8\ldots (\text{13 times}) \\ 20! &> 2\times3\times4\times5\times6\times 7\times (2^{3})^{13} \\ 20! &> 2^{40} \end{align*}
Comment: A similar problem was asked in mock test #6 (Problem B2).

(ii) True.
Let us prove the left hand side inequality. \begin{align*} f(x) &:= \ln x - 1 + 1/x\\ f^\prime(x) &= 1/x(1 - 1/x)\\ \end{align*} \(f^\prime < 0 \) for \(x\in(0,1)\) and \(>0\) for \(x\in(1,\infty)\). This means \(f\) is decreasing in the interval \( (0,1)\) and increasing in the interval \( (1,\infty) \). It reaches the minimum at \(x=1\) which is zero. Hence \(f(x)\geq 0\) for all \(x > 0\).
The right hand side inequality can be proved similarly. \begin{align*} g(x) &:= x - 1 - \ln x\\ g^\prime(x) &= 1 - 1/x \end{align*} Again, \( g(x) \) attains its minima of zero at \(x=1\).

Points on a sphere

B2, 2015

Let \(p, q\) and \(r\) be real numbers with \(p^{2}+q^{2}+r^{2}=1\)

(a) Prove the inequality \(3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3} \leq 2\)

(b) Also find the smallest possible value of \(3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3} \).

Specify exactly when the smallest and the largest possible value is achieved.


We have \begin{align} 3p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3}&=(q+r)\left(3 p^{2}+2 q^{2}+2 r^{2}-2 q r\right)\\ &=(q+r)\left(3\left(p^{2}+q^{2}+r^{2}\right)-\left(q^{2}+r^{2}+2 q r\right)\right)\\ &=(q+r)\left(3-(q+r)^{2}\right)\\ &=x\left(3-x^{2}\right)=3 x-x^{3} \end{align} where \(x=q+r.\)

Let us examine possible values of \(x\) in view of the constraint \(p^{2}+q^{2}+r^{2}=1\).

We have \(2 q r \leq q^{2}+r^{2}\) e.g. because \((q-r)^{2} \geq 0 .\)

Adding \(q^{2}+r^{2},\) we get \(q^{2}+r^{2}+2 q r \leq \) \(2 q^{2}+2 r^{2} \leq 2,\) because \(q^{2}+r^{2} \leq p^{2}+q^{2}+r^{2}=1\). Thus \((q+r)^{2} \leq 2\). So \(-\sqrt{2} \leq q+r \leq \sqrt{2}\)

Note that equalities are achieved precisely when \(p=0\) and \(q=r=\pm 1 / \sqrt{2}\).

Thus altogether we have to find extrema of the odd function \(f(x)=3 x-x^{3}\) over the interval \([-\sqrt{2}, \sqrt{2}] .\) The critical points are when \(f^{\prime}(x)=3-3 x^{2}=0,\) i.e. \(x=\pm 1 .\) Thus we need to see only \(f(\pm \sqrt{2})=\pm \sqrt{2}\) and \(f(\pm 1)=\pm 2 .\) Therefore \(-2 \leq 3 p^{2} q+3 p^{2} r+\) \(2 q^{3}+2 r^{3} \leq 2 .\) Moreover, \(3 p^{2} q+3 p^{2} r+2 q^{3}+2 r^{3}=\pm 2\) precisely when \(x=q+r=\pm 1\)

In each case, this gives a line segment in the \(q r\)-plane joining \( (\pm 1,0) \) and \((0,\pm 1) .\) Note that both these segments lie within the circle \(q^{2}+r^{2}=1,\) so each point on them leads to two valid points \((p, q, r)\) on the unit sphere.