Answer
$(2x+y)^5 = 32x^5 + 80x^4y +80x^3y^2 + 40x^2y^3 + 10xy^4+y^5$
Work Step by Step
$(2x+y)^5$
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + n(n-1)(n-2)(n-3)/4!* a^{n-3}*b^3+ n(n-1)(n-2)(n-3)(n-4)/5!*a^{n-4}*b^4+ n(n-1)(n-2)(n-3)(n-4)(n-5)/6!*a^{n-5}*b^5+ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)/6!*a^{n-6}*b^6$
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + n(n-1)(n-2)(n-3)/4!* a^{n-3}*b^3+ n(n-1)(n-2)(n-3)(n-4)/5!*a^{n-4}*b^4+ n(n-1)(n-2)(n-3)(n-4)(n-5)/6!*a^{n-5}*b^5$
$a=2x$
$b=y$
$n=5$
$(2x+y)^5 = (2x)^5 + 5/1!* (2x)^{5-1}*y^1 + 5(5-1)/2!* (2x)^{5-2}*y^2 + 5(5-1)(5-2)/3!* (2x)^{5-3}*y^3 + 5(5-1)(5-2)(5-3)/4!* (2x)^{5-4}*y^4+ 5(5-1)(5-2)(5-3)(5-4)/5!*(2x)^{5-5}*y^5$
$(2x+y)^5 = (2x)^5 + 5/1!* (2x)^{4}*y^1 + 5(4)/2!* (2x)^{3}*y^2 + 5(4)(3)/3!* (2x)^{2}*y^3 + 5(4)(3)(2)/4!* (2x)^{1}*y^4+ 5(4)(3)(2)(1)/5!*(2x)^{0}*y^5$
$(2x+y)^5 = (2x)^5 + 5*(2x)^{4}*y^1 + 5(2)*(2x)^{3}*y^2 + 5*2*(2x)^{2}*y^3 + 5*(2x)^{1}*y^4+ (2x)^{0}*y^5$
$(2x+y)^5 = (2x)^5 + 5*16x^4*y^1 + 5(2)*(2x)^{3}*y^2 + 5*2*(2x)^{2}*y^3 + 5*(2x)^{1}*y^4+ (2x)^{0}*y^5$
$(2x+y)^5 = 32x^5 + 80x^4y +10*8x^3y^2 + 10*4x^2y^3 + 10xy^4+y^5$
$(2x+y)^5 = 32x^5 + 80x^4y +80x^3y^2 + 40x^2y^3 + 10xy^4+y^5$
