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) $(-54,315)$
Show Answer
Answer:
Correct Answer: 2. (c)
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)
$\Rightarrow(105 \times 9)-45 a+b=0$
$\Rightarrow 45 a-b=945$
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)
$\Rightarrow-12285+945 a-45 b=0$
$\Rightarrow \quad 63 a-3 b=819$
$$ \Rightarrow \quad 21 a-b=273 $$
From Eqs. (i) and (ii), we get
$$ 24 a=672 \Rightarrow a=28 $$
So, $\quad b=315$
$\Rightarrow \quad(a, b)=(28,315)$
Key Idea Use the general term (or $(r+1)$ th term) in the expansion of binomial $(a+b)^{n}$
i.e.
$$ T _{r+1}={ }^{n} C _r a^{n-r} b^{r} $$
Let a binomial $2 x^{2}-{\frac{3}{x^{2}}}^{6}$, it’s $(r+1)$ th term
$$ \begin{aligned} & =T _{r+1}={ }^{6} C _r\left(2 x^{2}\right)^{6-r}-\frac{3}{x^{2}} \\ & ={ }^{6} C _r(-3)^{r}(2)^{6-r} x^{12-2 r-2 r} \\ & ={ }^{6} C _r(-3)^{r}(2)^{6-r} x^{12-4 r} \end{aligned} $$
Now, the term independent of $x$ in the expansion of $\frac{1}{60}-\frac{x^{8}}{81} \quad 2 x^{2}-\frac{3}{x^{2}}$
$=$ the term independent of $x$ in the expansion of $\frac{1}{60} 2 x^{2}-{\frac{3}{x^{2}}}^{6}+$ the term independent of $x$ in the expansion of $-\frac{x^{8}}{81} 2 x^{2}-{\frac{3}{x^{2}}}^{6}$
$\begin{array}{rrr}=\frac{{ }^{6} C _3}{60}(-3)^{3}(2)^{6-3} x^{12-4(3)} & \text { [put } r=3] \ +-\frac{1}{81} & { }^{6} C _5(-3)^{5}(2)^{6-5} x^{12-4(5)} x^{8} & \text { [put } r=5]\end{array}$
$=\frac{1}{3}(-3)^{3} 2^{3}+\frac{3^{5} \times 2(6)}{81}$
$=36-72=-36$
