Discrete Random Variable Quick Question! 5* for quick & right answer

Just need some quick help really:

The random variable X has the discrete uniform distribution:

P(X = x) = 1/n,
x = 1, 2, ..., n.

Given that E(X) = 5, show that n = 9.

E(X)=1/n+2/n+3/n+...+n/n
=(1+2+3+...+n)/n
=[n(n+1)/2]/n
=(n+1)/2 = 9 so you can now find n.

E(x) = (1/n) Σ i = 5
Σ i = 5n

1+2+3+4+5+6+7+8+9 = 5n
9(10)/2 = 5n
45 = 5n
n=9
--------------------------------------…
1+2+3+4+5+6+7+8+9 = sum of the first 9 natural numbers = 9(9+1)/2 = 9(10)/2 = 45

E(X)=(1/n) (1+2+...+n).
The sum of the integers from 1 to n is n(n+1)/2; so (1/n) n(n+1)/2 = 5.
(n+1)/2=5; n+1=10 and n=9.

NiggahzinParis