Question

If the maximum value of the term independent of $t$ in the expansion of ${\left(t^2{x}^{\frac{1}{5}}+\frac{(1-x{)}^{\frac{1}{10}}}{t}\right)}^{15},x\ge 0$, is $K$, then $8K$ is equal to ______.

Answer 1

Answer: 6006

${\left(t^2{x}^{\frac{1}{5}}+\frac{(1-x{)}^{\frac{1}{10}}}{t}\right)}^{15}$
$T_{r+1}={}^{15}{C}_r{\left(t^2{x}^{\frac{1}{5}}\right)}^{15-r}\cdot \frac{(1-x{)}^{\frac{r}{10}}}{t^r}$
For independent of $t$,
$30-2r-r=0$
$\Rightarrow r=10$
So, Maximum value of ${}^{15}{C}_{10}x(1-x)$ will be at
$x=\frac{1}{2}$
i.e. $6006$