Binomial Theorem 1 Question 3
4. The smallest natural number $n$, such that the coefficient of $x$ in the expansion of $x^{2}+\frac{1}{x^{3}} \quad{ }^{n}$ is ${ }^{n} C _{23}$, is (2019 Main, 10 April II)
(a) 35
(b) 23
(c) 58
(d) 38
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Answer:
Correct Answer: 4. (b)
Solution:
- Given binomial is $x^{2}+{\frac{1}{x^{3}}}^{n}$, its $(r+1)^{\text {th }}$ term, is
$$ \begin{aligned} T _{r+1} & ={ }^{n} C _r\left(x^{2}\right)^{n-r} \frac{1}{x^{3}}={ }^{r} C _r x^{2 n-2 r} \frac{1}{x^{3 r}} \\ & ={ }^{n} C _r x^{2 n-2 r-3 r}={ }^{n} C _r x^{2 n-5 r} \end{aligned} $$
For the coefficient of $x$,
$$ 2 n-5 r=1 \Rightarrow 2 n=5 r+1 \ldots \text { (i) } $$
As coefficient of $x$ is given as ${ }^{n} C _{23}$, then either $r=23$ or $n-r=23$.
If $r=23$, then from Eq. (i), we get
$2 n=5(23)+1$
$\Rightarrow 2 n=115+1 \Rightarrow 2 n=116 \Rightarrow n=58$.
If $n-r=23$, then from Eq. (i) on replacing the value of ${ }^{\text {c }}$ $r$, we get $2 n=5(n-23)+1$
$\Rightarrow \quad 2 n=5 n-115+1 \quad \Rightarrow \quad 3 n=114 \Rightarrow n=38$
So, the required smallest natural number $n=38$.
