$${t_n} = \frac{{n + 2}}{{n\left( {n + 1} \right)}} \cdot {\left( {\frac{1}{2}} \right)^n} = \frac{{2\left( {n + 1} \right) - n}}{{n\left( {n + 1} \right)}} \cdot {\left( {\frac{1}{2}} \right)^n} = \frac{1}{n} \cdot {\left( {\frac{1}{2}} \right)^{n - 1}} - \frac{1}{{n + 1}} \cdot {\left( {\frac{1}{2}} \right)^n}.$$

If $$n$$ is odd, the required sum is
$$\eqalign{ & {1^2} + {2.2^2} + {3^2} + {2.4^2} + ...... + 2.{\left( {n + 1} \right)^2} + {n^2} \cr & = \frac{{\left( {n - 1} \right){{\left( {n - 1 + 1} \right)}^2}}}{2} + {n^2} \cr & \left[ {\because \left( {n - 1} \right){\text{ is even}}} \right. \cr & \therefore {\text{ using given formula for the sum of }}\left( {n - 1} \right)\left. {{\text{terms}}{\text{.}}} \right] \cr & = \left( {\frac{{n - 1}}{2} + 1} \right){n^2} \cr & = \frac{{{n^2}\left( {n + 1} \right)}}{2} \cr} $$

$$\left( {2n + r} \right)r = {\left( {n + r} \right)^2} - {n^2} = \left\{ {1 + 3 + 5 + .....\,{\text{to }}\left( {n + r} \right){\text{terms}}} \right\} - \left\{ {1 + 3 + 5 + .....\,{\text{to }}n\,\,{\text{terms}}} \right\}.$$

$${\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right){\text{ and }}{\log _{10}}\left( {{2^x} + 3} \right){\text{ are in A}}{\text{.P}}{\text{.}}$$
Hence, common difference will be same.
$$\eqalign{ & \therefore {\text{lo}}{{\text{g}}_{10}}\left( {{2^x} - 1} \right) - {\log _{10}}2 \cr & = \log \left( {{2^x} + 3} \right) - {\log _{10}}\left( {{2^x} - 1} \right) \cr & \therefore {\log _{10}}\left( {\frac{{{2^x} - 1}}{2}} \right) = {\log _{10}}\left( {\frac{{{2^x} + 3}}{{{2^x} - 1}}} \right) \cr & \Rightarrow \frac{{{2^x} - 1}}{2} = \frac{{{2^x} + 3}}{{{2^x} - 1}} \cr & \Rightarrow {\left( {{2^x} - 1} \right)^2} = 2\left( {{2^x} + 3} \right) \cr & \Rightarrow {2^{2x}} - {2^{x + 1}} + 1 = {2^{x + 1}} + 6 \cr & \Rightarrow {2^{2x}} - {2^{x + 2}} = 5 \cr & {\text{Let }}{2^x} = y,{\text{ then}} \cr & {y^2} - 4y - 5 = 0 \cr & \Rightarrow {y^2} - 5y + y - 5 = 0 \cr & \Rightarrow y\left( {y - 5} \right) + 1\left( {y - 5} \right) = 0 \cr & y = - 1,y = 5 \cr & {\text{Therefore, }}{2^x} = 5 \cr & x = {\log _2}5. \cr} $$

$$\eqalign{ & {\left( {ap - b} \right)^2} + {\left( {bp - c} \right)^2} + {\left( {cp - d} \right)^2} \leqslant 0 \cr & \Rightarrow \,\,ap - b = 0,bp - c = 0,cp - d = 0 \cr & \therefore \,\,p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}. \cr} $$

$$\eqalign{ & AL_1^2 + {L_1}M_1^2 = \left( {{a^2} + {1^2}} \right) + \left\{ {{{\left( {a - 1} \right)}^2} + {1^2}} \right\} \cr & AL_2^2 + {L_2}M_2^2 = \left( {{a^2} + {2^2}} \right) + \left\{ {{{\left( {a - 2} \right)}^2} + {2^2}} \right\} \cr & ........................................................ \cr & AL_{a - 1}^2 + {L_{a - 1}}M_{a - 1}^2 = {a^2} + {\left( {a - 1} \right)^2} + \left\{ {{1^2} + {{\left( {a - 1} \right)}^2}} \right\}. \cr} $$
∴ the required sum
$$\eqalign{ & = \left( {a - 1} \right){a^2} + \left\{ {{1^2} + {2^2} + ..... + {{\left( {a - 1} \right)}^2}} \right\} + 2\left\{ {{1^2} + {2^2} + ..... + {{\left( {a - 1} \right)}^2}} \right\} \cr & = \left( {a - 1} \right){a^2} + 3 \cdot \frac{{\left( {a - 1} \right)a\left( {2a - 1} \right)}}{6} = a\left( {a - 1} \right)\left\{ {a + \frac{{2a - 1}}{2}} \right\}. \cr} $$

$$\eqalign{ & a{r^4} = 2 \cr & a \times ar \times a{r^2} \times a{r^3} \times a{r^4} \times a{r^5} \times a{r^6} \times a{r^7} \times a{r^8} \cr & = {a^9}{r^{36}} = {\left( {a{r^4}} \right)^9} = {2^9} = 512 \cr} $$

$$\eqalign{ & 2b = a + c;\,\,a,p,b,q,c\,\,{\text{are in A}}{\text{.P.}} \cr & {\text{Hence, }}p = \frac{{a + b}}{2}{\text{ and }}q = \frac{{b + c}}{2} \cr & {\text{Again, }}a,p',b,q'{\text{ and }}c = \,{\text{are in G}}{\text{.P.}} \cr & {\text{Hence, }}p' = \sqrt {ab} \,\,{\text{and }}q' = \sqrt {bc} \cr & {p^2} - {q^2} = \frac{{\left( {a - c} \right)\left( {a + c + 2b} \right)}}{4} \cr & = \frac{{\left( {a - c} \right)\left( {2b + 2b} \right)}}{4}\,\,\,\left[ {\because a + c = 2b} \right] \cr & = \left( {a - c} \right)b = ab - bc = p{'^2} - q{'^2} \cr} $$

$$\frac{x}{y},$$ e.t.c., are positive. $$A \geqslant G$$
$$ \Rightarrow \,\,\frac{{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}}{3} \geqslant \root 3 \of {\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}} = 1.$$

$$\frac{{\sin x + \cos x + {\text{cosec}}2x}}{3} \geqslant \root 3 \of {\sin x \cdot \cos x \cdot {\text{cosec}}2x} = \root 3 \of {\frac{1}{2}} .$$