Show that the power of $x$ with the largest coefficient in the polynomial $\left( 1 + \frac { 2 x } { 3 } \right) ^ { 20 }$ is 8 , i.e., if we write the given polynomial as $\sum _ { i } a _ { i } x ^ { i }$ then the largest coefficient $a _ { i }$ is $a _ { 8 }$.

Let $a _ { 0 } , a _ { 1 } , \ldots , a _ { 23 }$ be real numbers such that
$$\left( 1 + \frac { 2 } { 5 } x \right) ^ { 23 } = \sum _ { i = 0 } ^ { 23 } a _ { i } x ^ { i }$$
for every real number $x$. Let $a _ { r }$ be the largest among the numbers $a _ { j }$ for $0 \leq j \leq 23$. Then the value of $r$ is $\_\_\_\_$.

If the maximum value of the term independent of $t$ in the expansion of $\left( t ^ { 2 } x ^ { \frac { 1 } { 5 } } + \frac { 1 - x ^ { \frac { 1 } { 10 } } } { t } \right)^{10}$, $x \geq 0$, is $K$, then $8K$ is equal to $\_\_\_\_$.

Let for the $9 ^ { \text {th } }$ term in the binomial expansion of $( 3 + 6 x ) ^ { n }$, in the increasing powers of $6 x$, to be the greatest for $x = \frac { 3 } { 2 }$, the least value of $n$ is $n _ { 0 }$. If $k$ is the ratio of the coefficient of $x ^ { 6 }$ to the coefficient of $x ^ { 3 }$, then $k + n _ { 0 }$ is equal to $\_\_\_\_$ .

Consider a polynomial in $x$ and $y$
$$P = ( 3 x + 4 y + 1 ) ^ { 5 } .$$
Let us denote the coefficient of $x ^ { n } y$ in the expansion of $P$ by $a _ { n }$, where $n$ is an integer. Note that $x ^ { 0 } = y ^ { 0 } = 1$.
(1) Let us find the value of the coefficient $a _ { 1 }$. First, we note that
$$P = \{ ( 3 x + 1 ) + 4 y \} ^ { 5 }$$
and use the binomial theorem to expand $P$. Then $x y$ appears when we expand the term AB $( 3 x + 1 ) ^ { \text {C} } y$. Further, the coefficient for $x$ in the expansion of $( 3 x + 1 ) ^ { \text {C} }$ is DE. It follows that
$$a _ { 1 } = \mathbf { F G H } .$$
(2) The number of values which $n$ can take is $\square$ in all. Also, the value of $a _ { n }$ is maximized at $n =$ $\square$ J .

2. For ALL APPLICANTS.

For $k$ a positive integer, we define the polynomial $p _ { k } ( x )$ as
$$p _ { k } ( x ) = ( 1 + x ) \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 3 } \right) \times \cdots \times \left( 1 + x ^ { k } \right) = a _ { 0 } + a _ { 1 } x + \cdots + a _ { N } x ^ { N } ,$$
denoting the coefficients of $p _ { k } ( x )$ as $a _ { 0 } , \ldots , a _ { N }$.
(i) Write down the degree $N$ of $p _ { k } ( x )$ in terms of $k$.
(ii) By setting $x = 1$, or otherwise, explain why
$$a _ { \max } \geqslant \frac { 2 ^ { k } } { N + 1 }$$
where $a _ { \text {max } }$ denotes the largest of the coefficients $a _ { 0 } , \ldots , a _ { N }$.
(iii) Fix $i \geqslant 0$. Explain why the value of $a _ { i }$ eventually becomes constant as $k$ increases.
A student correctly calculates for $k = 6$ that $p _ { 6 } ( x )$ equals
$$\begin{aligned} & 1 + x + x ^ { 2 } + 2 x ^ { 3 } + 2 x ^ { 4 } + 3 x ^ { 5 } + 4 x ^ { 6 } + 4 x ^ { 7 } + 4 x ^ { 8 } + 5 x ^ { 9 } + 5 x ^ { 10 } + 5 x ^ { 11 } \\ & + 5 x ^ { 12 } + 4 x ^ { 13 } + 4 x ^ { 14 } + 4 x ^ { 15 } + 3 x ^ { 16 } + 2 x ^ { 17 } + 2 x ^ { 18 } + x ^ { 19 } + x ^ { 20 } + x ^ { 21 } \end{aligned}$$
(iv) On the basis of this calculation, the student guesses that
$$a _ { i } = a _ { N - i } \quad \text { for } \quad 0 \leqslant i \leqslant N$$
By substituting $x ^ { - 1 }$ for $x$, or otherwise, show that the student's guess is correct for all positive integers $k$.
(v) On the basis of the same calculation, the student guesses that all whole numbers in the range $1,2 , \ldots , a _ { \text {max } }$ appear amongst the coefficients $a _ { 0 } , \ldots , a _ { N }$, for all positive integers $k$.
Use part (ii) to show that in this case the student's guess is wrong. Justify your answer.
If you require additional space please use the pages at the end of the booklet