Explanation

(b)
The $n$th term of the sequence is
$\ldots n(n + 1)n\ldots 321$
$= 1 \times 10^{2n} + 2 \times 10^{2n - 1} + 3 \times 10^{2n - 2} + \ldots + n \times 10^{n + 1} + (n + 1) \times 10^{n}\ \times 10^{n - 1} + ... + 3 \times 10^{2} + 2 \times 10 + 1$
Let $S = 1 \times 10^{2n} + 2 \times 10^{2n - 1} + ... + n \times 10^{n} + 1$ (AG series)
$\frac{1}{10}\ S_{1} = 1 \times 10^{2n - 1} + ... + (n - 1) \times 10^{n - 1} + n \times 10^{n}$
On solving, we get
$\frac{9}{10}S_{1} = 1 \times 10^{2n} + ... + 1 \times 10^{n + 1} - n \times 10^{n}$
$S_{1} = \frac{10}{9}\left\lbrack 10^{2n} + 10^{2n - 1} + \ldots + 10^{n + 1} - n \times 10^{n} \right\rbrack$
$S_{1} = \frac{10}{9}\left\lbrack \frac{10^{2n}\left( 1 - \frac{1}{10^{n}} \right)}{1 - \frac{1}{10}} - n \times 10^{n} \right\rbrack$
$= \left( \frac{10}{9} \right)^{2}\left\lbrack 10^{2n}\left( 1 - \frac{1}{10^{n}} \right) \right\rbrack - \left\lbrack \left( \frac{10}{9} \right)n \times 10^{n} \right\rbrack$
Again, let
$S_{2} = (n + 1) \times 10^{n} + n \times 10^{n - 1} + ... + 2 \times 10 + 1$
$\frac{\frac{1}{10}S_{2} = (n + 1) \times 10^{n - 1} + ... + 3 \times 10 + 2 + \frac{1}{10}}{\frac{9}{10}S_{2} = (n + 1) \times 10^{n} - \left\lbrack 10^{n - 1} + 10^{n - 2} + ... + 1 + \frac{1}{10} \right\rbrack}$
$= (n + 1) \times 10^{n} - \frac{10^{n - 1}\left( 1 - \frac{1}{10^{n - 1}} \right)}{1 - \frac{1}{10}}$
$\Rightarrow \ S_{2} = \frac{10}{9}(n + 1) \times 10^{n} - \left( \frac{10}{9} \right)^{2}10^{n - 1}\left( 1 - \frac{1}{10^{n - 1}} \right)$
$\therefore\$The $n$th term = $S_{1} + S_{2}$
$= \left( \frac{10}{9} \right)^{2}\left\lbrack 10^{2n}\left( 1 - \frac{1}{10^{n - 1}} \right) - 10^{n - 1}\left( 1 - \frac{1}{10^{n - 1}} \right) \right\rbrack + \frac{10}{9}\left\lbrack - n \times 10^{n} + (n + 1) \times 10^{n} \right\rbrack$
$= \left( \frac{10}{9} \right)^{2}\left\lbrack 10^{2n} - {2.10}^{n - 1} + \frac{1}{10^{2}} \right\rbrack$
$= \frac{10}{9}\left\lbrack 10^{n} - \frac{1}{10} \right\rbrack^{2}$
$= \left\lbrack \frac{10^{n - 1} - 1}{9} \right\rbrack^{2}$
$= \left\lbrack \frac{999\ldots(n + 1)times}{9} \right\rbrack^{2}$
$= \left\lbrack 111\ldots(n + 1)times \right\rbrack^{2} = (odd\ number)^{2}$