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(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k

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$\frac{c}{4}$

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(B) The sum of probabilities is $\sum_{k=0}^{\infty} P(X=k) = 1$. Given $P(X=k) = \frac{3k+1}{2^{k+c}}$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$. Let $S = \sum_{k=0}^{\infty} (3k+1) x^k$ where $x = \frac{1}{2}$. $S = 3 \sum_{k=0}^{\infty} k x^k + \sum_{k=0}^{\infty} x^k = 3 \frac{x}{(1-x)^2} + \frac{1}{1-x}$. Substituting $x = \frac{1}{2}$: $S = 3 \frac{1/2}{(1/2)^2} + \frac{1}{1/2} = 3(2) + 2 = 8$. Thus,$\frac{1}{2^c} (8) = 1 \implies 2^3 = 2^c \implies c = 3$. We need to find $P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)$. $P(X=k) = \frac{3k+1}{2^{k+3}}$. $P(X=0) = \frac{1}{8}, P(X=1) = \frac{4}{16} = \frac{2}{8}, P(X=2) = \frac{7}{32}, P(X=3) = \frac{10}{64} = \frac{5}{32}$. Sum $= \frac{4}{32} + \frac{8}{32} + \frac{7}{32} + \frac{5}{32} = \frac{24}{32} = \frac{3}{4}$. Since $c=3$,the option $\frac{c}{4} = \frac{3}{4}$ matches option $B$.

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

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