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Question Number 209027 by Spillover last updated on 30/Jun/24

	Answered by Spillover last updated on 05/Jul/24

	$f (x) = \frac{1}{2^{k + 1}} x = 1, 2, 3, 4, 5, . . . . . n$ $f r o m M o m e n t g e n e r a t i n g f u n c t i o n (M G F)$ $M_{x} (t) = E (e^{t x})$ $M_{x} (t) = \underset{x = 1}{\sum^{n}} e^{t x} f (x)$ $M_{x} (t) = \underset{x = 1}{\sum^{n}} e^{t x} . \frac{1}{2^{k + 1}}$ $M_{x} (t) = \frac{1}{2^{k + 1}} \underset{x = 1}{\sum^{n}} e^{t x}$ $M_{x} (t) = \frac{1}{2^{k + 1}} (e^{t} + e^{2 t} + e^{3 t} + e^{4 t} . . . . . e^{n t})$ $T h i s i s g e o m e t r i c s e r i e s$ $f r o m \underset{x = 1}{\sum^{n}} {a r}^{n - 1} = \frac{a (1 - r^{n})}{1 - r}$ $a = e^{t} r = e^{t} \frac{e^{t} (1 - e^{n t})}{1 - r}$ $M_{x} (t) = \frac{1}{2^{k + 1}} [\frac{e^{t} (1 - e^{n t})}{1 - r}]$ $1^{s t} d e r i v a t i v e = m e a n$ $= M_{x} (t) = \frac{1}{2^{k + 1}} [\frac{e^{t} (1 - e^{n t})}{1 - e^{t}}]$ $M_{x}^{^{'}} (t) = \frac{1}{2^{k + 1}} [\frac{e^{t} (1 - e^{n t}) - e^{t n} e^{t}}{{(1 - e^{t})}^{2}}]$ $t = 0$ $M e a n (\bar{x}) = - \frac{1}{2^{k + 1}}$ $a p p l y q u o t i e n t r u l e$ $M_{x}^{^{'}} (t) = \frac{d}{d t} [\frac{1}{2^{k + 1}} \frac{e^{t} (1 - e^{n t}) - e^{t n} e^{t}}{{(1 - e^{t})}^{2}}]$ $M_{x}^{^{″}} (t) = \frac{d^{2}}{{d t}^{2}} [\frac{1}{2^{k + 1}} . \frac{e^{t} (1 - e^{n t}) - e^{t n} e^{t}}{{(1 - e^{t})}^{2}}]$
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