Question

Prove that coefficient of $x^4$ in (1 + x - $2x^2{)}^6$ is -45 and if complete expansion of the expression is $1+a_{1}x+a_2{x}^2+....+a_{1}2x^{12}$
Prove that $a_2+a_4+a_6+...+a_{12}=31$

Answer 1

${(1+x-2x^2)}^6=(1-x{)}^6(1+2x{)}^6$
$=(1-6x+15x^2-20x^3+15x^4......)\times [1+6(2x)+15(2x{)}^2+20(2x{)}^3+15(2x{)}^4+....]$
${}^6{C}_1=6,{}^6{C}_2=15,{}^6{C}_3=20,{}^6{C}_4=15$ etc.
Multiply and the term of $x^4$ will be
$[1\cdot 15\cdot 2^4-6\cdot 20\cdot 2^3+15\cdot 15\cdot 2^2-20\cdot 6\cdot 2+15\cdot 1]x^4$
Hence the coefficient of $x^4$ is
$240-960+900-240+15=-60+15=-45$
${(1+x-2x^2)}^6=1+a_{1}x+a_2{x}^2+..........+a_{12}{x}^{12}$
Putting, x = 1, we get
$0=1+a_1+a_2+....+a_{12}$
Putting, x = -1 we get
$64=1-a_1+a_2-....+a_{12}$
Adding (1) and (2) , we get
$64=2(1+a_2+a_4+....)$$\therefore a_2+a_4+a_6+......+a_{12}=31$