Given,
(t2x51+t(1−x)101)15
Now rth term is given by,
Tr+1=Cr15(t2x51)15−r⋅tr(1−x)10r
For independent of t,
2(15−r)−r=0
30−2r−r=0
⇒r=10
So, Maximum value of C1015x(1−x) will be at x=21(\text{as }\frac{dy}{dx}(x-{x}^{2})=1-2x&1-2x=0\Rightarrow x=\frac{1}{2})
=C101521×(21)=43003
So, K=43003 and 8K=6006