How do you find the third term of #(4x-2/x)^8#?

1 Answer
Aug 8, 2017

Third term is # 458752 x^4 #

Explanation:

Binomial theorem:
#(a + b)^n =(nC_0) a^nb^0+ (nC_1) a^(n-1)b^1 +(nC_2) a^(n-2)b^2 +...... (nC_n)b^n#
Here # a=4x ; b=-2/x , n= 8 #

We know #nC_r= (n!)/(r!(n-r)!) :. 8C_2=(8*7)/2=28#

#3# rd term is #T_3= 8 C_2 * (4x)^(8-2) * ( -2/x)^2# or

#T_3 = 28 * 4^6 * x^6 * 2^2 * 1/x^2 = 28* 4^7 * x^4# or

#T_3= 458752 x^4#

Third term is # 458752 x^4 # [Ans]