Question

What is the 4th term of the expanded binomial (2x – 3y)^6

Answer 1

The 4th term of the expanded binomial is
`-4320x^3y^3`

Explanation:

Considering:

`$ (x+y)^n = \sum_(k=0)^(n) \binom{n}{k} x^(n-k)y^k$`

`$ (2x-3y)^6 = \sum_(k=0)^(6) \binom{6}{k} (2x)^(6-k)(-3y)^k$`

Now, you gotta calculate for every value of
`k`

`$ (2x-3y)^6 = \binom{6}{0} (2x)^(6-0)(-3y)^0 + \binom{6}{1} (2x)^(6-1)(-3y)^1 + \binom{6}{2} (2x)^(6-2)(-3y)^2 + \\ $`

`$\binom{6}{3} (2x)^(6-3)(-3y)^3 + \binom{6}{4} (2x)^(6-4)(-3y)^4 + \binom{6}{5} (2x)^(6-5)(-3y)^5 + \binom{6}{6} (2x)^(6-6)(-3y)^6 $`

I will not write every product, but just solve following the steps:

For
`k=0`

`$\binom{6}{0} (2x)^(6-0)(-3y)^0$`

`$(6!)/((6-0)!(0!)) (2x)^(6-0)(-3y)^0$`

`$ (6!)/(6!) \left(2x\right)^(6-0)\cdot 1$`

`$1\cdot \:1\cdot \left(2x\right)^(6-0)$`

`$2^6x^6$`

`64x^6`

`(2x-3y)^6=64x^6-576x^5y+2160x^4y^2-4320x^3y^3+4860x^2y^4-2916xy^5+729y^6`