Solution:
(i) $\sum_{n=2}^{10} 4^{n}$
$=4^{2}+4^{3}+4^{4}+\ldots+4^{10}$
Where, $a=4^{2}=16, r=4^{3} / 4^{2}=4, n=9$
Using the formula,
The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
$=16\left(4^{9}-1\right) /(4-1)$
$\begin{array}{l}
=16\left(4^{9}-1\right) / 3 \\
=16 / 3\left[4^{9}-1\right]
\end{array}$