7.Binomial Theorem
normal

જો $\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$ ના દ્રીપદી વિસ્તરણમાં, $x^{10}$ નો સહગુણક $5^{ k } l$ હોય, જ્યાં $l, k \in N$ છે તથા $l$ અને $5$ પરસ્પર અવિભાજય છે,તો $k=\dots\dots$

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

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