- Home
- Standard 11
- Mathematics
7.Binomial Theorem
hard
Let $\alpha > 0$, be the smallest number such that the expansion of $\left(x^{\frac{2}{3}}+\frac{2}{x^3}\right)^{30}$ has a term $\beta x^{-\alpha}, \beta \in N$. Then $\alpha$ is equal to $.............$.
A
$2$
B
$4$
C
$6$
D
$8$
(JEE MAIN-2023)
Solution
$T _{ r +1}={ }^{30} C _{ r }\left( x ^{2 / 3}\right)^{30- r }\left(\frac{2}{ x ^3}\right)^{ r }$
$={ }^{30} C _{ r } \cdot 2^{ r } \cdot x ^{\frac{60-11 r }{3}}$
$\frac{60-11 r }{3} < 0 \Rightarrow 11 r > 60 \Rightarrow r >\frac{60}{11} \Rightarrow r =6$$T _7={ }^{30} C _6 \cdot 2^6 x ^{-2}$
We have also observed $\beta={ }^{30} C _6(2)^6$ is a natural number.
$\therefore \alpha=2$
Standard 11
Mathematics
