+23254 If you know combinations:
= C(5,0)(3b)5(0.5)0 + C(5,1)(3b)4(0.5)1 + C(5,2)(3b)3(0.5)2 + C(5,3)(3b)2(0.5)3
+ C(5,4)(3b)1(0.5)4 + C(5,5)(3b)0(0.5)5
= 1(3b)5(0.5)0 + 5(3b)4(0.5)1 + 10(3b)3(0.5)2 + 10(3b)2(0.5)3
+ 5(3b)1(0.5)4 + 1(3b)0(0.5)5
= 1(243b5)(1) + 5(81b4)(0.5) + 10(27b3)(0.25) + 10(9b2)(0.125)
+ 5(3b)(0.0625) + 1(1)(0.03125)
= 243b3 + 202.5b4 + 67.5b3 + 11.25b2 + 0.9375b + 0.03125
If you don't know the rule of combinations, use the 5th row of Pascal's triangle.
+118696 Sorry anon - That is not correct.
you need to use the binomial expansion
I am on my phone at the moment so it is all to difficult for me right now.
if no one gives a timely answer why not Google binomial expansion and watch a you tube presentation.
it would most likely be easier to understand anyway. :)
+23254 If you know combinations:
= C(5,0)(3b)5(0.5)0 + C(5,1)(3b)4(0.5)1 + C(5,2)(3b)3(0.5)2 + C(5,3)(3b)2(0.5)3
+ C(5,4)(3b)1(0.5)4 + C(5,5)(3b)0(0.5)5
= 1(3b)5(0.5)0 + 5(3b)4(0.5)1 + 10(3b)3(0.5)2 + 10(3b)2(0.5)3
+ 5(3b)1(0.5)4 + 1(3b)0(0.5)5
= 1(243b5)(1) + 5(81b4)(0.5) + 10(27b3)(0.25) + 10(9b2)(0.125)
+ 5(3b)(0.0625) + 1(1)(0.03125)
= 243b3 + 202.5b4 + 67.5b3 + 11.25b2 + 0.9375b + 0.03125
If you don't know the rule of combinations, use the 5th row of Pascal's triangle.
