Question

If the coefficient of x10 in the binomial expansion of
$\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}} + \frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$
is 5k.l, where l, k∈N and l is co-prime to 5, then k is equal to ___________.

Answer 1

The correct answer is 5
$T_{r+1} = ^{60}C_r(x^{\frac{1}{2}})^{60-r}(x^{-\frac{1}{3}})r(5^{\frac{-1}{4}})^{60-r}(5^{\frac{1}{2}})^r$
for
$x^{10} \frac{60-r}{2}-\frac{r}{3}=10$
$⇒ 180-3r-2r=60$
⇒ r = 24
∴ Coeff. of
$x^{10} = \frac{^{60}C_{24}}{5^9} 5^{12}$ = 5k I
as I and 5 are coprime
k = 3 + exponent of 5 in $^{60}C_{24}$
$= 3 + \left(\left\lfloor\frac{60}{5}\right\rfloor + \left\lfloor\frac{60}{5^2}\right\rfloor - \left\lfloor\frac{24}{5}\right\rfloor - \left\lfloor\frac{24}{5^2}\right\rfloor - \left\lfloor\frac{36}{5}\right\rfloor - \left\lfloor\frac{36}{5^2}\right\rfloor\right)$
= 3 + (12 + 2 – 4 – 0 – 7 – 1)
= 3 + 2 = 5