Term will be independent of t when 30−3r=0⇒r=10 ∴T10+1=T11 will be independent of t ∴T11=‌15C10⋅x‌
15−10
5
⋅(1−x)‌
10
10
=‌15C10⋅x1⋅(1−x)1 T11 will be maximum when x(1−x) is maximum. Let f(x)=x(1−x)=x−x2 f(x) is maximum or minimum when f′(x)=0 ∴f′(x)=1−2x For maximum ∕ minimum f′(x)=0 ∴1−2x=0 ⇒x=‌
1
2
Now, f′′(x)=−2<0 ∴ At x=‌
1
2
,f(x) maximum ∴ Maximum value of T11 is =‌15C10⋅‌