Question

can anyone help me understand this sum?
sum from k=0 to 6 of k(6 choose k)(1/6)^k(5/6)^(6-k)

Answer 1

i understand that if that first k was not there then you could use the binomial therom and get 1 but im not sure what to do becasue of that k

Answer 2

I also know wolfram alpha computes it as 1

Answer 3

i also get 1 if i compute i term by term but im looking for an explination that does not rely on that

Answer 4

Just checked via calculator:
sum(seq(k(6nCrk)(1/6)^k(5/6)^(6-k)) = 1

what do you mean you are not sure what to do with the k? You plug in 'n' the starting value.

Answer 5

right so if the k was not there you could represent the whole series as ((1/6)+(5/6))^6