UPSC NDA Math Model Paper 8

For the given expression (bx2+

1

bx

)13 , n = 13
Tr+1=‌13Cr(bx)13−r(

1

bx

)r
Tr+1=‌13Crb13−r(x)13−r(

1

brxr

)
Tr+1=‌13Crb13−2r(x)26−3r
Now, in order to find out coefficient of x8,26−3r must be 8.
i.e. 26−3r=8
r=6
Hence putting r=6 in equation (1) we get,
T6+1=‌13C6b13−12(x)26−18
T6+1=‌13C6bx8
Therefore of coefficient of x8 is equal to ‌13C6b