What is the correct expansion of the binomial (x+y)^5

Question

asked Jan 6, 2017 808 views

2 Answers

Alexey Podlasov · Answer 1 · 2017-01-11T16:23:35+0000

Answer: The correct expansion is,

x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Explanation:

Since, by the binomial expansion,

$(p+q)^n=\sum_(r=0)^(n) ^nC_r (p)^(n-r)(q)^r$

Where,

^nC_r=(n!)/(r!(n-r)!)

Here, p = x and q = y and n = 5,

By substituting values,

$(x+y)^5=\sum_(r=0)^(5) ^5C_r (x)^(n-r)(y)^r$

=^5C_0(x)^(5-0)(y)^0+^5C_1 (x)^(5-1)(y)^1+^5C_2 (x)^(5-2)(y)^2+^5C_3 (x)^(5-3)(y)^3+^5C_4 (x)^(5-4)(y)^4+^5C_5(x)^(5-5)(y)^(5)

=1(x)^5(y)^0+(5!)/(4!(5-4)!)x^4y^1+(5!)/(3!(5-3)!)x^3y^2+(5!)/(2!(5-2)!)x^2y^3+(5!)/(1!(5-1)!)xy^4+(5!)/(5!(5-5)!)x^0y^5

=x^5+(5!)/(4!1!)x^4y^1+(5!)/(3!2!)x^3y^2+(5!)/(2!3!)x^2y^3+(5!)/(1!4!)x^1y^4+(5!)/(5!0!)x^0y^5

=x^5+(5* 4!)/(4!)x^4y^1+(5* 4* 3!)/(3!2!)x^3y^2+(5* 4* 3!)/(2!3!)x^2y^3+(5* 4!)/(4!)x^1y^4+(5!)/(5!)y^5

=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5

Which is the required expansion.