Question

What is the coefficient of the x^5y^5-term in the binomial expansion of (2x -3y)10?

Answer 1

 a answer is (2x−5y)  and do the math 32x5−400x4y+2000x3y2−5000x2y3+6250xy4−3125y5

Answer 2

Answer:

The coefficient of  $x^5y^5$ is (-1959552).    

Step-by-step explanation:

Given : Expression $(2x-3y)^{10}$

To find : What is the coefficient of the term $x^5y^5$ in the binomial expansion of expression ?

Solution :

The binomial expansion is $(x+y)^n=\sum ^nC_k x^{n-k} y^k$

Where, $^nC_k=\frac{n!}{(n-k)!k!}$

On comparison with given expression  $(2x-3y)^{10}$

x=2x , y=-3y and n=10, k=0,1,.....,10.

Substituting in the formula and expand,

$(2x-3y)^{10}=\sum ^{10}C_k (2x)^{10-k} (-3y)^k$

$(2x-3y)^{10}=^{10}C_0(2x)^{10-0} (-3y)^0+^{10}C_1(2x)^{10-1} (-3y)^1+^{10}C_2(2x)^{10-2} (-3y)^2+^{10}C_3(2x)^{10-3} (-3y)^3+^{10}C_4(2x)^{10-4} (-3y)^4+^{10}C_5(2x)^{10-5} (-3y)^5+^{10}C_6(2x)^{10-6} (-3y)^6+^{10}C_7(2x)^{10-7} (-3y)^7+^{10}C_8(2x)^{10-8} (-3y)^8+^{10}C_9(2x)^{10-9} (-3y)^9+^{10}C_{10}(2x)^{10-10} (-3y)^{10]$

$(2x-3y)^{10}=1024x^{10}-15360x^9y+103680x^8y^2-414720x^7y^3+1088640x^6y^4-1959552x^5y^5+2449440x^4y^6-2099520x^3y^7+1180980x^2y^8-393660xy^9+59049y^{10}$

So, The coefficient of  $x^5y^5$ is (-1959552).