Question

Find the coefficient of $t^{32}$ in the expansion of ${(1+t^2)}^{12}(1+t^{12})(1+t^{24})$

Answer 1

${(1+t^2)}^{12}(1+t^{12})(1+t^{24})$

$=(1+t^{12})(1+t^{24}){(1+t^2)}^{12}$

$=(1+t^{12}+t^{24}+t^{36})$expansion of ${(1+t^2)}^{12}$

$=(1+t^{12}+t^{24}+t^{36})(1+{12}_{C_1}{t}^2+{12}_{C_2}{\left(t^2\right)}^2+{12}_{C_3}{\left(t^2\right)}^3+{12}_{C_4}{\left(t^2\right)}^4+{12}_{C_5}{\left(t^2\right)}^5+{12}_{C_6}{\left(t^2\right)}^6+{12}_{C_7}{\left(t^2\right)}^7+{12}_{C_8}{\left(t^2\right)}^8+{12}_{C_9}{\left(t^2\right)}^9+{12}_{C_{10}}{\left(t^2\right)}^{10}+{12}_{C_{11}}+{12}_{C_{12}}{\left(t^2\right)}^{12})$

$=(1+t^{12}+t^{24}+t^{36})(1+{12}_{C_1}{t}^2+{12}_{C_2}{t}^4+{12}_{C_3}{t}^6+{12}_{C_4}{t}^8+{12}_{C_5}{t}^{10}+{12}_{C_6}{t}^{10}+{12}_{C_7}{t}^{14}+{12}_{C_8}{t}^{16}+{12}_{C_9}{t}^{18}+{12}_{C_{10}}{t}^{20}+{12}_{C_{11}}{t}^{22}+{12}_{C_{12}}{t}^{24})$

Terms containing $t^{32}$ is

$=\left(t^{24}\right){12}_{C_4}{t}^8+\left(t^{12}\right){12}_{C_{10}}{t}^{20}$

$={12}_{C_4}+{12}_{C_{10}}$

$=\frac{12!}{8!4!}+\frac{12!}{10!2!}$

$=\frac{12\times 11\times 10\times 9\times 8!}{8!\times 4\times 3\times 2\times 1}+\frac{12\times 11\times 10!}{10!\times 2\times 1}$

$=\frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}+\frac{12\times 11}{2}$

$=11\times 5\times 9+6\times 11$

$=495+66$

$=561$ on simplification