If $$n$$ is odd, the required sum is

$${1^2} + {2.2^2} + {3^2} + {2.4^2} + ......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2.{\left( {n - 1} \right)^2} + {n^2}$$

$$ = {{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}} \over 2} + {n^2}$$

[ As $$\left( {n - 1} \right)$$ is even

$$\therefore$$ using given formula for the sum of $$\left( {n - 1} \right)$$ terms.]

$$ = \left( {{{n - 1} \over 2} + 1} \right){n^2} = {{{n^2}\left( {n + 1} \right)} \over 2}$$

$${T_m} = a + \left( {m - 1} \right)d = {1 \over n}...........\left( 1 \right)$$

$${T_n} = a + \left( {n - 1} \right)d = {1 \over m}..........\left( 2 \right)$$

$$\left( 1 \right) - \left( 2 \right) \Rightarrow \left( {m - n} \right)d$$

$$ = {1 \over n} - {1 \over m} \Rightarrow d = {1 \over {mn}}$$

From $$\left( 1 \right)$$ $$a = {1 \over {mn}} \Rightarrow a - d = 0$$

$${1 \over {1.2}} - {1 \over {2.3}} + {1 \over {3.4}}..........\infty $$

$$\left| {{T_n}} \right| = {1 \over {n\left( {n + 1} \right)}} = \left( {{1 \over n} - {1 \over {n + 1}}} \right)$$

$$S = {T_1} - {T_2} + {T_3} - {T_4} + {T_5}.........\infty $$

$$ = \left( {{1 \over 1} - {1 \over 2}} \right) - \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \left( {{1 \over 4} - {1 \over 5}} \right).......$$

$$ = 1 - 2\left[ {{1 \over 2} - {1 \over 3} + {1 \over 4} - {1 \over 5}.........\infty } \right]$$

$$ = 1 - 2\left[ { - \log \left( {1 + 1} \right) + 1} \right]$$

$$ = 2\log 2 - 1 = \log \left( {{4 \over e}} \right)$$

$${1^3} - {2^3} + {3^3} - {4^3} + ...... + {9^3}$$

$$ = {1^3} + {2^3} + {3^3} + ...... + {9^3}$$

$$\,\,\,\,\,\,\,\,\,\, - 2\left( {{2^3} + {4^3} + {6^3} + {8^3}} \right)$$

$$ = {\left[ {{{9 \times 10} \over 2}} \right]^2} - {2.2^3}\left[ {{1^3} + {2^3} + {3^3} + {4^3}} \right]$$

$$ = {\left( {45} \right)^2} - 16.{\left[ {{{4 \times 5} \over 2}} \right]^2}$$

$$ = 2025 - 1600 = 425$$