(1/2)+(2/4)+(3/8)+(6/(16))+((11)/(32))+((20)/(64))+((37)/(128))+... =?


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Question Number 136489 by liberty last updated on 22/Mar/21

$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{6}{16} + \frac{11}{32} + \frac{20}{64} + \frac{37}{128} + . . . = ?$
	Answered by Olaf last updated on 22/Mar/21
	$Tribonacci numbers : T_0 = 0, T_1 = 1, T_2 = 0 T_n = T_(n−1) +T_(n−2) +T_(n−3) ⇒ T_n = {0, 1, 0, 1, 2, 3, 6, 11, 20, 37...} S = Σ_(n=3) ^∞ (T_n /2^(n−2) ) = 4Σ_(n=3) ^∞ (T_n /2^n ) = 4Σ_(n=0) ^∞ (T_n /2^n )−2 The generator function of {T_n } is given by : F(X) = ((X−X^2 )/(1−X−X^2 −X^3 )) = Σ_(n=0) ^∞ T_n X^n ⇒ S = 4F((1/2))−2 S = 4((1/4)/(1−(1/2)−(1/4)−(1/8)))−2 = 6$
	$Tribonacci numbers :$ $T_{0} = 0, T_{1} = 1, T_{2} = 0$ $T_{n} = T_{n - 1} + T_{n - 2} + T_{n - 3}$ $\Rightarrow T_{n} = {0, 1, 0, 1, 2, 3, 6, 11, 20, 37 . . .}$ $S = \underset{n = 3}{\sum^{\infty}} \frac{T_{n}}{2^{n - 2}} = 4 \underset{n = 3}{\sum^{\infty}} \frac{T_{n}}{2^{n}} = 4 \underset{n = 0}{\sum^{\infty}} \frac{T_{n}}{2^{n}} - 2$ $The generator function of {T_{n}} is$ $given by :$ $F (X) = \frac{X - X^{2}}{1 - X - X^{2} - X^{3}} = \underset{n = 0}{\sum^{\infty}} T_{n} X^{n}$ $\Rightarrow S = 4 F (\frac{1}{2}) - 2$ $S = 4 \frac{\frac{1}{4}}{1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{8}} - 2 = 6$
	Commented by Olaf last updated on 22/Mar/21

	$a t l e a s t, I t r i e d : - ($
	Commented by liberty last updated on 23/Mar/21

	$s o o r y s i r . a n s w e r 6 i s c o r r e c t .$
	Commented by liberty last updated on 23/Mar/21

	$t h a n k s m u c h$
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