f.4 maths binomial expansion

1 General term=(nCr)a^(n-r)b^r

For (4x-1/(2x) )^10,general term=(10Cr)[(4x)^(10-r)](-1/2x)^r
=(10Cr)[(4)^(10-r)](-1/2)^r[x^(10-2r)]

So,when r=5,we have the constant term and

constant term
= 10C5 (4)^5 [-1/2]^5
= -8064

2 (1+4x)^n
coefficient of x^2 = nC2 (4^2) = 16 nC2
coefficient of x = nC1 (4) = 4n
16 nC2 = (10) 4 n
16 n(n-1)/2 = 40n
4(n-1) = 20
n = 6

1. General Term = (10Ck) (4x)^(10-k) (-1/2)^k (x)^-k
= (10Ck) (4)^(10-k) (-1/2)^k (x)^(10-2k)
For constant term,
10 - 2k = 0
k = 5
Constant term = (10C5) 4^5 (-1/2)^5
= -8064

2. General Term = (nCk)4^k (x)^k

For coeff. of x^2, k = 2
For coeff. of x, k = 1

[(nC1)4^1] x 10 = (nC2) 4^2
40n = [n(n-1)/2] x 16
8n^2 - 48n= 0
n = 6

(4x-1/(2x) )^10
constant term
= 10C5 (4x)^5 [-1/(2x)]^5
= (252) (4^5) (-1/2^5)
= 8064
------------------------
(1+4x)^n
coefficient of x^2 = nC2 (4^2) = 16 nC2
coefficient of x = nC1 (4) = 4n
16 nC2 = (10) 4 n
16 n(n-1)/2 = 40n
4(n-1) = 20
n = 6