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7.Binomial Theorem
normal
If the coefficient of $x ^{10}$ in the binomial expansion of $\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$ is $5^{ k } l$, where $l, k \in N$ and $l$ is coprime to $5$ , then $k$ is equal to
A
$5$
B
$6$
C
$7$
D
$8$
(JEE MAIN-2022)
Solution
$\left(\frac{\sqrt{x}}{5^{1 / 4}}+\frac{\sqrt{5}}{x^{1 / 3}}\right)^{60}$
$T_{r+1}={ }^{60} C_{r}\left(\frac{x^{1 / 2}}{5^{1 / 4}}\right)^{60-r}\left(\frac{5^{1 / 2}}{x^{1 / 3}}\right) r$
$={ }^{60} C_{r} 5 \frac{3 r-60}{4} x \frac{180-5 r}{6}$
$\frac{180-5 r}{6}=10 \Rightarrow r=24$
Coeff. of $x^{10}={ }^{60} C_{24} 5^{3}=\frac{\mid 60}{|24| 36} 5^{3}$
Powers of $5$ in $={ }^{60} C_{24} \cdot 5^{3}=\frac{5^{14}}{5^{4} \times 5^{8}} \times 5^{3}=5^{5}$
Standard 11
Mathematics
