If the p.m.f of a d.r.v. X is P(X=x) `= {{:((x)/(n(n+1))", for","x = 0,1,2,3,...,n"),(" 0,"," otherwise"):}` then E(X) = ……

Answer

To find the expected value \( E(X) \) of the discrete random variable \( X \) with the given probability mass function (p.m.f.):

\[ P(X = x) = \frac{x}{n(n+1)} \quad \text{for } x = 0, 1, 2, \ldots, n \]
\[ P(X = x) = 0 \quad \text{otherwise} \]

### Step 1: Write the formula for expected value
The expected value \( E(X) \) is calculated using the formula:
\[ E(X) = \sum_{x=0}^{n} x \cdot P(X = x) \]

### Step 2: Substitute the p.m.f. into the expected value formula
Substituting the p.m.f. into the expected value formula gives:
\[ E(X) = \sum_{x=0}^{n} x \cdot \frac{x}{n(n+1)} \]

### Step 3: Simplify the expression
This can be simplified as:
\[ E(X) = \frac{1}{n(n+1)} \sum_{x=0}^{n} x^2 \]

### Step 4: Use the formula for the sum of squares
The sum of squares of the first \( n \) natural numbers is given by:
\[ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6} \]
Note that \( x = 0 \) contributes \( 0^2 = 0 \) to the sum, so we can start from \( 1 \).

### Step 5: Substitute the sum of squares into the expected value
Substituting this result into the expression for \( E(X) \):
\[ E(X) = \frac{1}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6} \]

### Step 6: Simplify the expression
The \( n(n+1) \) terms cancel out:
\[ E(X) = \frac{2n + 1}{6} \]

### Final Result
Thus, the expected value \( E(X) \) is:
\[ E(X) = \frac{2n + 1}{6} \]

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