Question

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

Answer 1

Answer: -1959552

Explanation:

The (r+1) th term of binomial expansion
`(a+b)^n` is

`T_(r+1)=^nC_r\ a^(n-r)\ (b)^r`

Give expression :
`(2x-3y)^(10)`

here , a= 2x , b= -3y and n= 10

Comparing
`x^5y^5` to
`a^(n-r)\ (b)^r` , we get

r=5

The (5+1)th term will be :
`^(10)C_5\ (2x)^(10-5)\ (-3y)^5`

`(10!)/(5!(10-5)!)\ (2x)^(5)\ (-3y)^5`

`252\ (32x^5)\ (-243y^5)=252*32*-243x^5y^5=-1959552x^5y^5`

Hence, the coefficient of
`x^5y^5` =-1959552

Answer 2

If we have to expand the expression

`(a - b)^n=_(0)^(n)\textrm{C} a^n - _(1)^(n)\textrm{C} a^(n-1) b+_(2)^(n)\textrm{C}a^(n-2)b^2-_(3)^(n)\textrm{C}a^(n-3)b^3+...............................\pm _(n)^(n)\textrm{C} b^n`

Whether the sign before last term is positive or negative, Depends on whether , n is even or odd.

Let,
`T_(r+1)` be the term containing ,
`x^5 y^5` in the binomial expansion of
`(2x-3y)^(10)`.

`T_(r+1)=_(r)^(10)\textrm{C} (2x)^r* (-3y)^(10-r)`

If we replace r, by 5 in the above equation we get coefficient of ,
`x^5 y^5` in the binomial expansion of
`(2x-3y)^(10)`.

`T_(5+1)=_(5)^(10)\textrm{C} (2x)^5* (-3y)^(10-5)\\\\ T_(6)=_(5)^(10)\textrm{C}2^5* (-3)^5* x^5 y^5`

So, Coefficient of
`x^5 y^5`

`=_(5)^(10)\textrm{C}2^5* (-3)^5\\\\= (10!)/((10-5)!* 5!)* 32 * (-243)\\\\=252 * 32 * (-243)\\\\= -1959552`