Question

If $(1+ax+b{x}^2)(1-2x{)}^{18}$ be expressed in ascending powers of x such that co-efficients of $x^3$ and $x^4$ are zero, then (a, b) is equal to

• A. $(16,\frac{272}{3})$
• B. $(14,\frac{272}{3})$
• C. $(16,\frac{251}{3})$
• D. $(14,\frac{251}{3})$

Answer 1

Answer: (A)

Since co-eff. of $x^4=0=$ co-eff. of $x^3$ in $(1+ax+b{x}^2)(1-2x{)}^{18}$
$\therefore {}^{18}{c}_4{\left(-2\right)}^4+a{\cdot }^{18}{c}_3{\left(-2\right)}^3+b{\cdot }^{18}{c}_2{\left(-2\right)}^2=0$
and ${}^{18}{c}_3{\left(-2\right)}^3+a{\cdot }^{18}{c}_2{\left(-2\right)}^2+b{\cdot }^{18}{c}_1\left(-2\right)=0$
$\Rightarrow 32a-3b=240$ and $51a-3b=544$
$\Rightarrow a=16$, $b=\frac{272}{3}$
$\therefore \left(a,b\right)=\left(16,\frac{272}{3}\right)$.