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Q.
If $(1 + ax + bx^2) (1 - 2x)^{18}$ 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
Binomial Theorem
Solution:
Since co-eff. of $x^4 = 0 =$ co-eff. of $x^3$ in $(1 + ax + bx^2) \,(1 - 2x)^{18}$
$\therefore \,{}^{18}c_{4}\left(-2\right)^{4}+a\cdot^{18}c_{3}\left(- 2\right)^{3}+b\cdot^{18}c_{2}\left(-2\right)^{2} = 0$
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 $51a - 3b = 544$
$\Rightarrow a = 16$, $b = \frac{272}{3}$
$\therefore \left(a, b\right) = \left(16, \frac{272}{3}\right)$.