Question

If ${(1+x+x^2)}^{12}=a_0+a_{1}x+a_2{x}^2+.....+a_{24}{x}^{24}$ where the value of $a_0+a_2+a_4+.....+a_{22}$ is $\frac{{27}^k-1}{2}$, then write the value of 'k'

Answer 1

$(1+x+x^2{)}^{12}=a_0+a_{1}x+a_2{x}^{2}e---a_{24}{x}^{24}$

Let $x=-1$,

$(1+(-1)+(-1{)}^2{)}^{12}=a_0-a_1+a_2----a_{24}----\left(1\right)$

$x=1,(1+1+1{)}^{12}=a_0+a_1+a_2-----a_{24}-----(2)$

on adding $\left(1\right)$ & $\left(2\right)$

$1^{12}+3^{12}=2(a_0+a_2+---a_{24})$

$\frac{3^{12}+1}{2}-1=a_0+a_2+---a_{22}$

$\frac{(3^3{)}^4-1}{2}=a_0+a_3+----a_{22}$

$\Rightarrow a_0+a_2+---a_{22}$

$\Rightarrow a_0+a_2+---a_{22}=\frac{(27{)}^4-1}{2}$

$k=4$