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 .