Binomial Theorem 1 Question 18
19. Coefficient of $x^{11}$ in the expansion of $\left(1+x^{2}\right)^{4}\left(1+x^{3}\right)^{7}\left(1+x^{4}\right)^{12}$ is
(2014 Adv.)
(a) 1051
(b) 1106
(c) 1113
(d) 1120
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Answer:
Correct Answer: 19. $(n=6)$
Solution:
- Coefficient of $x^{r}$ in $(1+x)^{n}$ is ${ }^{n} C _r$.
In this type of questions, we find different composition of terms where product will give us $x^{11}$.
Now, consider the following cases for $x^{11}$ in
$\left(1+x^{2}\right)^{4}\left(1+x^{3}\right)^{7}\left(1+x^{4}\right)^{12}$.
Coefficient of $x^{0} x^{3} x^{8}$; Coefficient of $x^{2} x^{9} x^{0}$
Coefficient of $x^{4} x^{3} x^{4}$; Coefficient of $x^{8} x^{3} x^{0}$
$={ }^{4} C _0 \times{ }^{7} C _1 \times{ }^{12} C _2+{ }^{4} C _1 \times{ }^{7} C _3 \times{ }^{12} C _0+{ }^{4} C _2 \times{ }^{7} C _1$
$=462+140+504+7=1113$
$\times{ }^{12} C _1+{ }^{4} C _4 \times{ }^{7} C _1 \times{ }^{12} C _0$
$\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-{\frac{(x-1)}{x-x^{1 / 2}}}^{10}$
$=\left(x^{1 / 3}+1\right)-\frac{(\sqrt{x}+1)}{\sqrt{x}}^{10}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$
$\therefore$ The general term is
$$ T _{r+1}={ }^{10} C _r\left(x^{1 / 3}\right)^{10-r}\left(-x^{-1 / 2}\right)^{r}={ }^{10} C _r(-1)^{r} x^{\frac{10-r}{3}-\frac{r}{2}} $$
For independent of $x$, put
$$ \begin{array}{rlrl} & & \frac{10-r}{3}-\frac{r}{2} & =0 \quad \Rightarrow \quad 20-2 r-3 r=0 \\ \Rightarrow & 20 & =5 r \Rightarrow r=4 \\ \therefore & T _5={ }^{10} C _4 & =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210 \end{array} $$
