Question

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

Answer 1

Answer:

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

Step-by-step 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 $k=0$

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

$\frac{6!}{(6-0)!(0!)} (2x)^{6-0}(-3y)^0$

$\frac{6!}{6!} \left(2x\right)^{6-0}\cdot 1$

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

$2^6x^6$

$64x^6$

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