Answer
(a) ${b_1} = {a_2} = \frac{1}{3}$, ${b_2} = {a_3} = \frac{1}{5}$, ${b_3} = {a_4} = \frac{1}{7}$
(b) ${c_1} = {a_4} = \frac{1}{7}$, ${c_2} = {a_5} = \frac{1}{9}$, ${c_3} = {a_6} = \frac{1}{{11}}$
(c) ${d_1} = {a_1}^2 = 1$, ${d_2} = {a_2}^2 = \frac{1}{9}$, ${d_3} = {a_3}^2 = \frac{1}{{25}}$
(d) ${e_1} = 2{a_1} - {a_2} = \frac{5}{3}$, ${e_2} = 2{a_2} - {a_3} = \frac{7}{{15}}$, ${e_3} = 2{a_3} - {a_4} = \frac{9}{{35}}$
Work Step by Step
(a) ${a_{n + 1}}$ for $n=1,2,3$:
${a_2} = \frac{1}{{2\cdot2 - 1}} = \frac{1}{3}$, ${a_3} = \frac{1}{{2\cdot3 - 1}} = \frac{1}{5}$, ${a_4} = \frac{1}{{2\cdot4 - 1}} = \frac{1}{7}$.
Since ${b_n} = {a_{n + 1}}$, so
${b_1} = {a_2} = \frac{1}{3}$, ${b_2} = {a_3} = \frac{1}{5}$, ${b_3} = {a_4} = \frac{1}{7}$.
(b) ${a_{n + 3}}$ for $n=1,2,3$:
${a_4} = \frac{1}{{2\cdot4 - 1}} = \frac{1}{7}$, ${a_5} = \frac{1}{{2\cdot5 - 1}} = \frac{1}{9}$, ${a_6} = \frac{1}{{2\cdot6 - 1}} = \frac{1}{{11}}$.
Since ${c_n} = {a_{n + 3}}$, so
${c_1} = {a_4} = \frac{1}{7}$, ${c_2} = {a_5} = \frac{1}{9}$, ${c_3} = {a_6} = \frac{1}{{11}}$.
(c) ${a_n}^2$ for $n=1,2,3$:
${a_1}^2 = {\left( {\frac{1}{{2\cdot1 - 1}}} \right)^2} = 1$, ${a_2}^2 = {\left( {\frac{1}{{2\cdot2 - 1}}} \right)^2} = \frac{1}{9}$, ${a_3}^2 = {\left( {\frac{1}{{2\cdot3 - 1}}} \right)^2} = \frac{1}{{25}}$.
Since ${d_n} = {a_n}^2$, so
${d_1} = {a_1}^2 = 1$, ${d_2} = {a_2}^2 = \frac{1}{9}$, ${d_3} = {a_3}^2 = \frac{1}{{25}}$.
(d) $2{a_n} - {a_{n + 1}}$ for $n=1,2,3$:
$2{a_1} - {a_2} = 2\left( {\frac{1}{{2\cdot1 - 1}}} \right) - \left( {\frac{1}{{2\cdot2 - 1}}} \right) = 2 - \frac{1}{3} = \frac{5}{3}$,
$2{a_2} - {a_3} = 2\left( {\frac{1}{{2\cdot2 - 1}}} \right) - \left( {\frac{1}{{2\cdot3 - 1}}} \right) = \frac{2}{3} - \frac{1}{5} = \frac{7}{{15}}$,
$2{a_3} - {a_4} = 2\left( {\frac{1}{{2\cdot3 - 1}}} \right) - \left( {\frac{1}{{2\cdot4 - 1}}} \right) = \frac{2}{5} - \frac{1}{7} = \frac{9}{{35}}$.
Since ${e_n} = 2{a_n} - {a_{n + 1}}$, so
${e_1} = 2{a_1} - {a_2} = \frac{5}{3}$, ${e_2} = 2{a_2} - {a_3} = \frac{7}{{15}}$, ${e_3} = 2{a_3} - {a_4} = \frac{9}{{35}}$.
