A random variable X has its range {0,1,2,3,………}. If `P(X=r)=(c(r+1))/(2^(r))` for r=0,1,2,……….. Then c=

Options

A. `1//2`
B. `1//3`
C. `1//4`
D. `1//5`

Answer

To find the value of \( c \) in the probability function \( P(X = r) = \frac{c(r+1)}{2^r} \) for \( r = 0, 1, 2, \ldots \), we need to ensure that the total probability sums to 1. This means we need to evaluate the sum:

\[ \sum_{r=0}^{\infty} P(X = r) = 1 \]

Substituting the expression for \( P(X = r) \):

\[ \sum_{r=0}^{\infty} \frac{c(r+1)}{2^r} = 1 \]

Now, we can factor out \( c \):

\[ c \sum_{r=0}^{\infty} \frac{(r+1)}{2^r} = 1 \]

Next, we need to evaluate the sum \( \sum_{r=0}^{\infty} \frac{(r+1)}{2^r} \). This can be split into two parts:

\[ \sum_{r=0}^{\infty} \frac{(r+1)}{2^r} = \sum_{r=0}^{\infty} \frac{r}{2^r} + \sum_{r=0}^{\infty} \frac{1}{2^r} \]

The second sum is a geometric series:

\[ \sum_{r=0}^{\infty} \frac{1}{2^r} = \frac{1}{1 - \frac{1}{2}} = 2 \]

Now, we need to calculate \( \sum_{r=0}^{\infty} \frac{r}{2^r} \). This can be found using the formula for the sum of a series:

\[ \sum_{r=0}^{\infty} \frac{r}{x^r} = \frac{x}{(x-1)^2} \quad \text{for } |x| > 1 \]

Setting \( x = 2 \):

\[ \sum_{r=0}^{\infty} \frac{r}{2^r} = \frac{2}{(2-1)^2} = 2 \]

Now we can combine both parts:

\[ \sum_{r=0}^{\infty} \frac{(r+1)}{2^r} = 2 + 2 = 4 \]

Substituting this back into our equation gives:

\[ c \cdot 4 = 1 \]

Solving for \( c \):

\[ c = \frac{1}{4} \]

Thus, the value of \( c \) is:

\[ \boxed{\frac{1}{4}} \]