Verify that f(x) = 2x/( k(k + 1)) for x = 1, 2, 3, . . . ,k can serve as the probability distribution of a random variable in a given range

If f(x) = 2x/k(k+1), x= 1, 2, 3,
4,....k.

To verify whether this is probability distribution
function, we should prove that {Sum f(x) over x = 1, 2, 3, 4,...k} =
1.

x takes a discrete value. That is x takes values like
 x= 1, 2, 3, 4, ......, k.

Let X be the random
variable

f(X =x) = 2x/k(k+1) = 2x/k(k+1), x= 1, 2, 3,
4....k.

Total frequency = {Sum f(X = x) , x= 1, 2, 3,
...n.} =
2*1/k(k+1)+2*2/k(k+1)+2*3/k9k+1)...2*k/k(k+1).

{Sum f(X =
x) , x= 1, 2, 3, ...n.} =
{2/k(k+1)}{1+2+3+..........k}

{Sum f(X = x) , x= 1, 2, 3,
...n.} = {2/k(k+1)}{Sum of the k natural numbers starting from
1}.

{Sum f(X = x) , x= 1, 2, 3, ...n.} =
{2/(k(k+1)}{k(+1)/2}

{Sum f(X = x) , x= 1, 2, 3, ...n.} =
1..

Therefore f(x) is a frequency density function. Or f(x)
is a probability distribution function.