Question

desenvolva , empregando a formula do binômio. (2x+1)elevado a 5

Answer 1

$(2x+1)^5 =$

$(\frac{5}{0}).(2x)^5.1^0+(\frac{5}{1}).(2x)^4.1^1+(\frac{5}{2}).(2x)^3.1^2+(\frac{5}{3}).(2x)^2.1^3+ (\frac{5}{4}).(2x)^1.1^4\\ +(\frac{5}{5}).(2x)^0.1^5 =$

$(2x)^5+(5).(2x)^4+(\frac{5}{2}).(2x)^3+(\frac{5}{3}).(2x)^2+ (\frac{5}{4}).(2x)+1 =$

$32x^5+5.16x^4+(\frac{5}{2}).8x^3+(\frac{5}{3}).4x^2+ (\frac{5}{4}).(2x)+1 =$

$32x^5+5.16x^4+ (\frac{5!}{2!.(5-2)!}).8x^3+(\frac{5!}{3!.(5-3)!}).4x^2+ (\frac{5!}{4!.(5-4)!}).(2x)+1 =$

$32x^5+5.16x^4+ (\frac{5.4}{2}).8x^3+(\frac{5.4}{2}).4x^2+ 5.(2x)+1 =$

$32x^5+80x^4+ 80x^3+40x^2+10x+1 =$