If the coefficients of x^3 and x^4 in the expansion of (1+ax+bx^2)(1-2x)^18 in powers of x…

Question

If the coefficients of $x^3$ and $x^4$ in the expansion of $(1+ax+bx^2)(1-2x)^{18}$ in powers of $x$ are both zero, then $(a,b)$ is equal to:

Accepted Answer

$$\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}$$

= $${\left( {1 - 2x} \right)^{18}} + ax{\left( {1 - 2x} \right)^{18}} + b{x^2}{\left( {1 - 2x} \right)^{18}}$$

= $$\left( {1 + ax + b{x^2}} \right)\left[ {{}^{18}{C_0} - {}^{18}{C_1}\left( {2x} \right) + {}^{18}{C_2}{{\left( {2x} \right)}^2} - {}^{18}{C_3}{{\left( {2x} \right)}^3} + ....} \right]$$

Coefficient of x3 is

$${\left( { - 2} \right)^3}.{}^{18}{C_3} + a{\left( { - 2} \right)^2}.{}^{18}{C_2} + b\left( { - 2} \right).{}^{18}{C_1}$$ = 0

$$ \Rightarrow 153a - 9b = 1632$$ ....... (1)

Coefficient of x4 is

$${\left( { - 2} \right)^4}.{}^{18}{C_4} + a{\left( { - 2} \right)^3}.{}^{18}{C_3} + b{\left( { - 2} \right)^2}.{}^{18}{C_2}$$ = 0

$$ \Rightarrow 3b - 32a = -240$$ ....... (2)

Solving (1) and (2), we get $a$ = 16, b = $${{172} \over 3}$$

Solutions · 1