Question

Coefficient of $t^{24}$ in ${(1-t)}^{12}{(t+t^2)}^{12}+(t+t^{12})(1-t^{24})$

Answer 1

${(1-t)}^{12}{(t+t^2)}^{12}+(t+t^{12})(1-t^{24})={(1-t)}^{12}{t}^{12}{(1+t)}^{12}+(t+t^{12})(1-t^{24})={(1-t^2)}^{12}{t}^{12}+t+t^{12}-t^{25}-t^{36}$

So in $1st$ term only $t^{24}$ would occur

$={(1-t^2)}^{12}{t}^{12}$

on$={(1-t^2)}^{12}$ co-efficient of $=t^{12}$

is required as $=t^{12}$ as multiplied in each term

$={(1-t^2)}^{12}$

$=12C_0{1}^{12}{\left({-t}^2\right)}^0+12C_1{\left(1\right)}^{11}{(-t^2)}^1+12C_2{(-t^2)}^2+12C_6{(-t^2)}^6+12C_{12}{(-t^2)}^{12}$

$=\frac{121}{6161}=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6}{6\times 5\times 4\times 3\times 2\times 1\times 6}=11\times 2\times 3\times 2\times 7=77\times 12=924$

Answer