If the coefficients of

Question:

If the coefficients of $x^{2}$ and $x^{3}$ are both zero, in the expansion of the expression $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ in powers of $x$, then the ordered pair $(a, b)$ is equal to :

  1. $(28,315)$

  2. $(-54,315)$

  3. $(-21,714)$

  4. $(24,861)$


Correct Option: 1

Solution:

Coefiicient of $\mathrm{x}^{2}={ }^{15} \mathrm{C}_{2} \times 9-3 \mathrm{a}\left({ }^{15} \mathrm{C}_{1}\right)+\mathrm{b}=0$

$\Rightarrow-45 \mathrm{a}+\mathrm{b}+{ }^{15} \mathrm{C}_{2} \times 9=0$ ..............(i)

Also, $-27 \times{ }^{15} \mathrm{C}_{3}+9 \mathrm{a} \times{ }^{15} \mathrm{C}_{2}-3 \mathrm{~b} \times{ }^{15} \mathrm{C}_{1}=0$

$\Rightarrow 9 \times{ }^{15} C_{2} a-45 b-27 \times{ }^{15} C_{3}=0$

$\Rightarrow 21 a-b-273=0$  .......(ii)

(i) $+$ (ii)

$-24 a+672=0$

$\Rightarrow a=28$

So, $b=315$

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