Binomial Theorem 1 Question 2
2.
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
(2019 Main, 10 April I)
(a) $(28,315)$
(b) $(-21,714)$
(c) $(28,861)$
(d) $(-54,315)$
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Answer:
Correct Answer: 2. (a)
Solution:
- Given expression is $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$. In the expansion of binomial $(1-3 x)^{15}$, the $(r+1)$ th term is
$ T_{r+1}={ }^{15} C_{r}(-3 x)^{r}={ }^{15} C_{r}(-3)^{r} x^{r} $
Now, coefficient of $x^{2}$, in the expansion of $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ is
${ }^{15} C_{2}(-3)^{2}+a^{15} C_{1}(-3)^{1}+b{ }^{15} C_{0}(-3)^{0}=0$ (given)
$\Rightarrow(105 \times 9)-45 a+b=0$
$\Rightarrow 45 a-b=945$ $\quad$ …….(i)
Similarly, the coefficient of $x^{3}$, in the expansion of $\left(1+a x+b x^{2}\right)(1-3 x)^{15}$ is
${ }^{15} C_{3}(-3)^{3}+a{ }^{15} C_{2}(-3)^{2}+b{ }^{15} C_{1}(-3)^{1}=0 \quad$ (given)
$\Rightarrow-12285+945 a-45 b=0$
$\Rightarrow \quad 63 a-3 b=819$
$ \Rightarrow \quad 21 a-b=273 $ $\quad$ …….(ii)
From Eqs. (i) and (ii), we get
$ 24 a=672 \Rightarrow a=28 $
So, $\quad b=315$
$\Rightarrow \quad(a, b)=(28,315)$
