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.
