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

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Answer 1 · 2017-08-08T12:10:23

Third term is $458752 x^4$

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]

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