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Q.
When $(3 x+5)^{100}$ is expanded, the largest power of 2 dividing the coefficient of $x^{39}$ is
Binomial Theorem
Solution:
$T _{ r +1}= { }^{100} C _{ r }(3 x )^{100- r } \cdot 5^{ r } $
$={ }^{100} C _{ r } \cdot 3^{100- r } \cdot 5^{ r } \cdot x ^{100- r }$
$\therefore 100- r =39 \Rightarrow r =61 $
$\therefore \text { Coefficient of } x ^{39}={ }^{100} C _{39} \cdot 3^{39} \cdot 5^{61} .$
Now ${ }^{100} C _{39}=\frac{(100) !}{39 !(61) !}=\frac{50+25+12+6+3+1}{(19+9+4+2+1)+(30+15+7+3+1)}=\frac{97}{(35)+(56)}=\frac{97}{91}=6$.
Hence largest power of 2 that divides the coefficient of $x^{39}$ is 6