Question

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

Answer 1

`(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5`

`(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6`

Explanation:

In order to find the values of
`(x+y)^5` and
`(x+y)^6`, you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

`(x+y)^5` =
`\sum _(i=0)^5\binom{5}{i}x^(\left(5-i\right))y^i` =
`(5!)/(0!\left(5-0\right)!)x^5y^0+(5!)/(1!\left(5-1\right)!)x^4y^1+(5!)/(2!\left(5-2\right)!)x^3y^2+(5!)/(3!\left(5-3\right)!)x^2y^3+(5!)/(4!\left(5-4\right)!)x^1y^4+(5!)/(5!\left(5-5\right)!)x^0y^5` =
`x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5`.

`(x+y)^6` =
`\sum _(i=0)^6\binom{6}{i}x^(\left(6-i\right))y^i` =
`(6!)/(0!\left(6-0\right)!)x^6y^0+(6!)/(1!\left(6-1\right)!)x^5y^1+(6!)/(2!\left(6-2\right)!)x^4y^2+(6!)/(3!\left(6-3\right)!)x^3y^3+(6!)/(4!\left(6-4\right)!)x^2y^4+(6!)/(5!\left(6-5\right)!)x^1y^5+(6!)/(6!\left(6-6\right)!)x^0y^6`=
`x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6`.