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(D) The sum of all probabilities must be $1$,so $\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$. Let $S = \sum_{x=0}^{\infty} (x+1)3^{-x} = 1 + \frac{2}{3} + \

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$\frac{1}{9}$

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(D) The sum of all probabilities must be $1$,so $\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$. Let $S = \sum_{x=0}^{\infty} (x+1)3^{-x} = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \ldots$. Then $\frac{1}{3}S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \ldots$. Subtracting the two equations: $S - \frac{1}{3}S = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots$. This is an infinite geometric series with $a=1$ and $r=\frac{1}{3}$,so $\frac{2}{3}S = \frac{1}{1 - 1/3} = \frac{3}{2}$. Thus $S = \frac{9}{4}$. Since $kS = 1$,we have $k = \frac{4}{9}$. We need to find $P(X \geq 3) = 1 - P(X=0) - P(X=1) - P(X=2)$. $P(X=0) = k(1)3^0 = k = \frac{4}{9}$. $P(X=1) = k(2)3^{-1} = \frac{2k}{3} = \frac{2}{3} 	imes \frac{4}{9} = \frac{8}{27}$. $P(X=2) = k(3)3^{-2} = \frac{3k}{9} = \frac{k}{3} = \frac{4}{27}$. $P(X \geq 3) = 1 - (\frac{4}{9} + \frac{8}{27} + \frac{4}{27}) = 1 - (\frac{12+8+4}{27}) = 1 - \frac{24}{27} = 1 - \frac{8}{9} = \frac{1}{9}$.

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

If the probability that the random variable $X$ takes the value $x$ is given by $P(X=x) = k(x+1)3^{-x}$,for $x = 0, 1, 2, 3, \ldots$,where $k$ is a constant,then $P(X \geq 3)$ is equal to

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