The smallest natural number n,

Question:

The smallest natural number $\mathrm{n}$, such that the coefficient of $x$ in the expansion of $\left(x^{2}+\frac{1}{x^{3}}\right)^{n}$

is ${ }^{\mathrm{n}} \mathrm{C}_{23}$, is :

  1. 35

  2. 38

  3. 23

  4. 58


Correct Option: , 2

Solution:

$\mathrm{T}_{\mathrm{r}}=\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{2 \mathrm{n}-2 \mathrm{r}} \cdot \mathrm{x}^{-3 \mathrm{r}}$

$2 n-5 r=1 \Rightarrow 2 n=5 r+1$

for $\mathrm{r}=15 . \mathrm{n}=38$

smallest value of $\mathrm{n}$ is 38 .

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