Question

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

Answer 1

This is the answer to your problem.

Answer 2

Answer: The correct expansion is,

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

Step-by-step explanation:

Since, by the binomial expansion,

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

Where,

$^nC_r=\frac{n!}{r!(n-r)!}$

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

By substituting values,

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

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

$=1(x)^5(y)^0+\frac{5!}{4!(5-4)!}x^4y^1+\frac{5!}{3!(5-3)!}x^3y^2+\frac{5!}{2!(5-2)!}x^2y^3+\frac{5!}{1!(5-1)!}xy^4+\frac{5!}{5!(5-5)!}x^0y^5$

$=x^5+\frac{5!}{4!1!}x^4y^1+\frac{5!}{3!2!}x^3y^2+\frac{5!}{2!3!}x^2y^3+\frac{5!}{1!4!}x^1y^4+\frac{5!}{5!0!}x^0y^5$

$=x^5+\frac{5\times 4!}{4!}x^4y^1+\frac{5\times 4\times 3!}{3!2!}x^3y^2+\frac{5\times 4\times 3!}{2!3!}x^2y^3+\frac{5\times 4!}{4!}x^1y^4+\frac{5!}{5!}y^5$

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

Which is the required expansion.