Question

When $(3x+5{)}^{100}$ is expanded, the largest power of 2 dividing the coefficient of $x^{39}$ is

• A. 6
• B. 7
• C. $2^6$
• D. $2^7$

Answer 1

Answer: (A)

$T_{r+1}={}^{100}{C}_r(3x{)}^{100-r}\cdot 5^r$
$={}^{100}{C}_r\cdot 3^{100-r}\cdot 5^r\cdot x^{100-r}$
$\therefore 100-r=39\Rightarrow r=61$
$\therefore \text{ Coefficient of }{x}^{39}={}^{100}{C}_{39}\cdot 3^{39}\cdot 5^{61}.$
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