Σ_(n=1) ^∞ (n^3 /3^n )=?


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Question Number 78460 by Tony Lin last updated on 17/Jan/20

$\underset{n = 1}{\sum^{\infty}} \frac{n^{3}}{3^{n}} = ?$
Commented by mathmax by abdo last updated on 18/Jan/20
$let s =Σ_(n=1) ^∞ (n^3 /3^n ) and f(x)=Σ_(n=0) ^∞ x^n with ∣x∣<1 ⇒f(x)=(1/(1−x)) ⇒f^((1)) (x)=(1/((1−x)^2 )) =Σ_(n=1) ^∞ nx^(n−1) ⇒(x/((x−1)^2 )) =Σ_(n=1) ^∞ nx^n ⇒ Σ_(n=1) ^∞ n^2 x^(n−1) =((x/((x−1)^2 )))^((1)) =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) =((x+1)/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n^2 x^n =((x^2 +x)/((1−x)^3 )) ⇒Σ_(n=1) ^∞ n^3 x^(n−1) =(((x^2 +x)/((1−x)^3 )))^((1)) =(((2x+1)(1−x)^3 +3(1−x)^2 (x^2 +x))/((1−x)^6 )) =(((2x+1)(1−x)+3(x^2 +x))/((1−x)^4 )) =((2x−2x^2 +1−x +3x^2 +3x)/((1−x)^4 )) =((x^2 +4x+1)/((1−x)^4 )) ⇒Σ_(n=1) ^∞ n^3 x^n =((x^3 +4x^2 +x)/((1−x)^4 )) ⇒s =f((1/3)) =((((1/3))^3 +4((1/3))^2 +(1/3))/((1−(1/3))^4 )) =(((1/3^3 )+(4/3^2 )+(1/3))/(2^4 /3^4 )) =((3^4 {(1/3^3 )+(4/3^2 )+(1/3)})/(16)) =((3+36 +27)/(16)) =((39+27)/(16)) =((66)/(16)) =((33)/8) ★ s=((33)/8)★$
$l e t s = \sum_{n = 1}^{\infty} \frac{n^{3}}{3^{n}} a n d f (x) = \sum_{n = 0}^{\infty} x^{n} w i t h ∣ x ∣< 1 \Rightarrow f (x) = \frac{1}{1 - x}$ $\Rightarrow f^{(1)} (x) = \frac{1}{{(1 - x)}^{2}} = \sum_{n = 1}^{\infty} {n x}^{n - 1} \Rightarrow \frac{x}{{(x - 1)}^{2}} = \sum_{n = 1}^{\infty} {n x}^{n} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} x^{n - 1} = {(\frac{x}{{(x - 1)}^{2}})}^{(1)} = \frac{{(x - 1)}^{2} - 2 (x - 1) x}{{(x - 1)}^{4}}$ $= \frac{x - 1 - 2 x}{{(x - 1)}^{3}} = \frac{- x - 1}{{(x - 1)}^{3}} = \frac{x + 1}{{(1 - x)}^{3}} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} x^{n} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow \sum_{n = 1}^{\infty} n^{3} x^{n - 1} = {(\frac{x^{2} + x}{{(1 - x)}^{3}})}^{(1)}$ $= \frac{(2 x + 1) {(1 - x)}^{3} + 3 {(1 - x)}^{2} (x^{2} + x)}{{(1 - x)}^{6}} = \frac{(2 x + 1) (1 - x) + 3 (x^{2} + x)}{{(1 - x)}^{4}}$ $= \frac{2 x - 2 x^{2} + 1 - x + 3 x^{2} + 3 x}{{(1 - x)}^{4}} = \frac{x^{2} + 4 x + 1}{{(1 - x)}^{4}} \Rightarrow \sum_{n = 1}^{\infty} n^{3} x^{n} = \frac{x^{3} + 4 x^{2} + x}{{(1 - x)}^{4}}$ $\Rightarrow s = f (\frac{1}{3}) = \frac{{(\frac{1}{3})}^{3} + 4 {(\frac{1}{3})}^{2} + \frac{1}{3}}{{(1 - \frac{1}{3})}^{4}} = \frac{\frac{1}{3^{3}} + \frac{4}{3^{2}} + \frac{1}{3}}{\frac{2^{4}}{3^{4}}}$ $= \frac{3^{4} {\frac{1}{3^{3}} + \frac{4}{3^{2}} + \frac{1}{3}}}{16} = \frac{3 + 36 + 27}{16} = \frac{39 + 27}{16} = \frac{66}{16} = \frac{33}{8}$ $★ s = \frac{33}{8} ★$
Commented by Tony Lin last updated on 18/Jan/20

$t h a n k s s i r$
Commented by mathmax by abdo last updated on 18/Jan/20

$y o u a r e w e l c o m e$
	Answered by mr W last updated on 17/Jan/20

	$1 + x + x^{2} + x^{3} + . . . = \frac{1}{1 - x} w i t h ∣ x ∣< 1$ $1 + 2 x + 3 x^{2} + 4 x^{3} + . . . = \frac{1}{{(1 - x)}^{2}}$ $x + 2 x^{2} + 3 x^{3} + 4 x^{4} + . . . = \frac{x}{{(1 - x)}^{2}}$ $1 + 2^{2} x + 3^{2} x^{2} + 4^{2} x^{3} + . . . = \frac{1}{{(1 - x)}^{2}} + \frac{2 x}{{(1 - x)}^{3}}$ $x + 2^{2} x^{2} + 3^{2} x^{3} + 4^{2} x^{4} + . . . = \frac{x}{{(1 - x)}^{2}} + \frac{2 x^{2}}{{(1 - x)}^{3}}$ $1 + 2^{3} x + 3^{3} x^{2} + 4^{3} x^{3} + . . . = \frac{1}{{(1 - x)}^{2}} + \frac{6 x}{{(1 - x)}^{3}} + \frac{6 x^{2}}{{(1 - x)}^{4}}$ $x + 2^{3} x^{2} + 3^{3} x^{3} + 4^{3} x^{4} + . . . = \frac{x}{{(1 - x)}^{2}} + \frac{6 x^{2}}{{(1 - x)}^{3}} + \frac{6 x^{3}}{{(1 - x)}^{4}}$ $l e t x = \frac{1}{3} :$ $\frac{1^{3}}{3} + \frac{2^{3}}{3^{2}} + \frac{3^{3}}{3^{3}} + \frac{4^{3}}{3^{4}} + . . . = \underset{n = 1}{\sum^{\infty}} \frac{n^{3}}{3^{n}} = \frac{1}{3} \times \frac{3^{2}}{2^{2}} + \frac{6}{3^{2}} \times \frac{3^{3}}{2^{3}} + \frac{6}{3^{3}} \times \frac{3^{4}}{2^{4}}$ $\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{n^{3}}{3^{n}} = \frac{33}{8}$
	Commented by Tony Lin last updated on 18/Jan/20

	$t h a n k s s i r$
	Answered by mind is power last updated on 17/Jan/20

	$x^{n} . n^{3}$ $n (n - 1) (n - 2) = n^{3} - {3 n}^{2} + 2 n$ $n^{3} = n (n - 1) (n - 2) + n (3 n - 2) = n (n - 1) (n - 2) + 3 n (n - 1) + n$ $\Rightarrow \sum_{n ⩾ 1} n (n - 1) (n - 2) x^{n} + 3 \sum_{n ⩾ 1} n (n - 1) x^{n} + \sum_{n ⩾ 1} {nx}^{n}$ $= x^{3} \sum_{n ⩾ 2} n (n - 1) (n - 2) x^{n - 3} + {3 x}^{2} \sum_{n ⩾ 2} n (n - 1) x^{n - 2} + x \sum_{n ⩾ 1} {nx}^{n - 1}$ $\sum_{n ⩾ 0} x^{n} = \frac{1}{1 - x}$ $\Rightarrow \sum_{n ⩾ 1} {nx}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow Σ n (n - 1) (n - 2) x^{n - 3} = \frac{6}{{(1 - x)}^{4}}$ $Σ n (n - 1) x^{n - 2} = \frac{2}{{(1 - x)}^{3}}$ $= \frac{{6 x}^{3}}{{(x - 1)}^{4}} + \frac{{6 x}^{2}}{{(1 - x)}^{3}} + \frac{x}{{(1 - x)}^{2}}$ $x = 3 \Rightarrow \frac{6.27}{16} + \frac{6.9}{- 8} + \frac{3}{4} = \frac{162 - 108 + 12}{16} = \frac{66}{16} = \frac{33}{8}$
	Commented by Tony Lin last updated on 18/Jan/20

	$t h a n k s s i r$
	Commented by mind is power last updated on 18/Jan/20

	$y^{'} re welcom$
	Answered by john santu last updated on 18/Jan/20

	$l e t S = \frac{1}{3} + \frac{8}{9} + \frac{27}{27} + \frac{64}{81} + \frac{125}{243} + . . . (1)$ $b y m u l t i p l y i n b o t h s i d e w i t h \frac{1}{3}$ $\frac{1}{3} S = \frac{1}{9} + \frac{8}{27} + \frac{27}{64} + \frac{64}{243} + . . . (2)$ $s u b s t r a c t e q u a t i o n s 1 a n d 2$ $\frac{2}{3} S = \frac{1}{3} + \frac{7}{9} + \frac{19}{27} + \frac{37}{81} + \frac{61}{243} + . . . (3)$ $\frac{2}{9} S = \frac{1}{9} + \frac{7}{27} + \frac{19}{81} + \frac{37}{243} + . . .$ $\frac{4}{9} S = \frac{1}{3} + \frac{6}{9} + \frac{12}{27} + \frac{18}{81} + \frac{24}{243} + . . .$ $\frac{4}{9} S = \frac{1}{3} + 6 (\frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \frac{4}{243} + . . .)$ $l e t P = \frac{1}{9} + \frac{2}{27} + \frac{3}{81} + \frac{4}{243} + . . .$ $\frac{1}{3} P = \frac{1}{27} + \frac{2}{81} + \frac{3}{243} + . . .$ $\frac{2}{3} P = \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + . . . = \frac{1}{6}$ $\Rightarrow P = \frac{1}{4} . n o w w e g e t$ $\frac{4}{9} S = \frac{1}{3} + 6 \times \frac{1}{4} \Rightarrow S = \frac{33}{8} . ★$
	Commented by Tony Lin last updated on 18/Jan/20

	$t h a n k s s i r$
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