M mrsir New member Joined Feb 14, 2013 Messages 2 Feb 14, 2013 #1 show that the following equation is true for all values of a and b [(a-b)+1]^5=a^5-5a^4(b-1)+10a^3(b-1)^2-10^2(b-1)^3=5a(b-1)^4-(b-1)^5
show that the following equation is true for all values of a and b [(a-b)+1]^5=a^5-5a^4(b-1)+10a^3(b-1)^2-10^2(b-1)^3=5a(b-1)^4-(b-1)^5
J JeffM Elite Member Joined Sep 14, 2012 Messages 7,874 Feb 15, 2013 #2 mrsir said: show that the following equation is true for all values of a and b [(a-b)+1]^5=a^5-5a^4(b-1)+10a^3(b-1)^2-10^2(b-1)^3=5a(b-1)^4-(b-1)^5 Click to expand... \(\displaystyle c = - b + 1 \implies \{(a - b) + 1\}^5 = \{a + (- b + 1\}^5 = (a + c)^5.\) Now apply the binomial theorem. Replace c with (- b + 1).
mrsir said: show that the following equation is true for all values of a and b [(a-b)+1]^5=a^5-5a^4(b-1)+10a^3(b-1)^2-10^2(b-1)^3=5a(b-1)^4-(b-1)^5 Click to expand... \(\displaystyle c = - b + 1 \implies \{(a - b) + 1\}^5 = \{a + (- b + 1\}^5 = (a + c)^5.\) Now apply the binomial theorem. Replace c with (- b + 1).
