Explanation

(d)
We have,
(a) $0.5 + 0.55 + 0.555 + ... = \frac{5}{9}\lbrack 0.9 + 0.99 + 0.999 + ...\rbrack$
$= \frac{5}{9}\lbrack(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + ...$to $n$ terms]
$= \frac{5}{9}\lbrack(1 + 1 + ...$to $n$ terms)$- \left. \ \left( \frac{1}{10} + \frac{1}{10^{2}} + \frac{1}{10^{3}} + ... + to\ n\ term \right) \right\rbrack$
$= \frac{5}{9}\left\lbrack n - \frac{\frac{1}{10}\left\{ 1 - \frac{1}{10^{n}} \right\}}{1 - \frac{1}{10}} \right\rbrack$
$= \frac{5n}{9} - \frac{5}{81}\left( 1 - 10^{- n} \right)$
(b) $8 + 88 + 888 + ...$to $n$ terms
$= \frac{8}{9}\lbrack 9 + 99 + 999 + ...$to $n$ terms]
$= \frac{8}{9}\lbrack(10 - 1) + \left( 10^{2} - 1 \right) + \left( 10^{3} - 1 \right) + ...$to $n$ terms]
$= \frac{8}{9}\lbrack(10 + 10^{2} + 10^{3} + \ldots + to\ n$ terms)$- (1 + 1 + 1 +$…+to $n$ terms)]
$= \frac{8}{9}\left\lbrack \frac{10\left( 10^{n} - 1 \right)}{10 - 1} - n \right\rbrack$
$= \frac{80}{81}\left( 10^{n} - 1 \right) - \frac{8n}{9}$
(c) the $n$th terms in the sequence is
$x_{n} = 1^{2} + 2^{2} + 3^{2} + ... + n^{2}$
$= \frac{n(n + 1)(2n + 1)}{6}$
$= \frac{1}{3}n^{3} + \frac{1}{2}n^{2} + \frac{1}{6}n$
$\therefore\$ The required sum $= \sum_{}^{}x_{n}$
$= \frac{1}{3}\sum_{}^{}{n^{3} + \frac{1}{2}}\sum_{}^{}{n^{2} + \frac{1}{6}}\sum_{}^{}n$
$= \frac{1}{3}\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2} + \frac{1}{2}.\frac{n(n + 1)(2n + 1)}{6} + \frac{1}{6}.\frac{n(n + 1)}{2}$
$= \frac{n(n + 1)}{12}\lbrack(n + 1) + (2n + 1) + 1$
$= \frac{n(n + 1)}{2}\left\lbrack n^{2} + 3n + 2 \right\rbrack$
$= \frac{n(n + 1)}{2}(n + 1)(n + 2)$
$= \frac{n(n + 1)^{2}(n + 2)}{12}$