Question

Find the middle term in the expansion of (2/x – x^2)^6.​

Answer 1

The middle term is
`-160x^(3)`

Explanation:

Given

`((2)/(x) - x^2)^6`

Required

Determine the middle term

This can be expanded using binomial expansion

i.e.

`(a + b)^n = a^n + ^nC_1a^(n-1)b^1 + ^nC_2a^(n-2)b^2 +....+b^n`

When n = 6;

`(a + b)^6 = a^6 + ^6C_1a^(6-1)b^1 + ^6C_2a^(6-2)b^2 + ^6C_3a^(6-3)b^3+ ^6C_4a^(6-4)b^4+ ^6C_5a^(6-5)b^5+b^n`

`(a + b)^6 = a^6 + ^6C_1a^(5)b^1 + ^6C_2a^(4)b^2 + ^6C_3a^(3)b^3+ ^6C_4a^(2)b^4+ ^6C_5a^(1)b^5+b^6`

`(a + b)^6 = a^6 + ^6C_1a^(5)b + ^6C_2a^(4)b^2 + ^6C_3a^(3)b^3+ ^6C_4a^(2)b^4+ ^6C_5ab^5+b^6`

The middle term here is given as:

`(a + b)^6 = ^6C_3a^(3)b^3`

In
`((2)/(x) - x^2)^6`

`a = (2)/(x)` and
`b = -x^2`

This gives:

`((2)/(x) - x^2)^6 = ^6C_3((2)/(x))^(3)(-x^2)^3`

`((2)/(x) - x^2)^6 = 20*((2)/(x))^(3)(-x^2)^3`

Solve all exponents

`((2)/(x) - x^2)^6 = 20*(8)/(x^3)*(-x^6)`

`((2)/(x) - x^2)^6 = -(20*8*x^6)/(x^3)`

`((2)/(x) - x^2)^6 = -160*x^(6-3)`

`((2)/(x) - x^2)^6 = -160*x^(3)`

`((2)/(x) - x^2)^6 = -160x^(3)`

Hence, the middle term is
`-160x^(3)`