Answer
$(p+(-2r))^5 = p^5 -10p^4r+40p^34r^2-80p^2r^3+80pr^4-32r^5$
Work Step by Step
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + . . . + b^n$
$a=p$
$b=-2r$
$n=5$
$(p+(-2r))^5 = p^5 + 5/1!* p^{5-1}*(-2r)^1 + 5(5-1)/2!* p^{5-2}*(-2r)^2 + 5(5-1)(5-2)/3!* p^{5-3}*(-2r)^3 + 5(5-1)(5-2)(5-3)/4!* p^{5-4}*(-2r)^4 + 5(5-1)(5-2)(5-3)(5-4)/5!* p^{5-5}*(-2r)^5$
$(p+(-2r))^5 = p^5 + 5*p^{4}*-2r+5(4)/2!* p^{3}*(-2r)^2 + 5(4)(3)/3!* p^{2}*(-2r)^3 + 5(4)(3)(2)/4!*p^{1}*(-2r)^4 + 5(4)(3)(2)(1)/5!*p^{0}*(-2r)^5$
$(p+(-2r))^5 = p^5 + 5p^4*-2r+5*2*p^3*4r^2 + 5(4)(3)/6*p^2*-8r^3 + 5*p*(-2r)^4 + (-2r)^5$
$(p+(-2r))^5 = p^5 -10p^4*r+40p^34r^2 -5*2*p^2*8r^3+5p*16r^4- 32r^5$
$(p+(-2r))^5 = p^5 -10p^4r+40p^34r^2-80p^2r^3+80pr^4-32r^5$
