







nC0 + nC1 + nC2 + .... + nCn = 2^n 
20180219 

From bristal: (QQ) Prove, nC0 + nC1 + nC2 + .... + nCn = 2^n. Answered by Penny Nom. 





0.999 ^ (500) 
20100307 

From debra: I just need to know how to solve the following problem without using a calculator: .999 ^ (500). I know the answer is .606, I just want to do it by hand since I can't use a calculator on my test. Answered by Penny Nom and Claude Tardif. 





Binomial coefficients 
20000321 

From Howard Lutz: How do you find each successive numerical term in this equation y+dy=(x+dx)^{5} =x^{5}+5*x^{4}dx+10*x^{3}(dx)^{2}+10*x^^{2}(dx)^{3}+5*x(dx)^{4}+(dx)^{5} I would appreciate an explanation of the method to find the numeric coefficient in a binomial expansion Answered by Penny Nom. 

