Σ_(n≥0) n^2 x^n


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Question Number 61724 by aliesam last updated on 07/Jun/19

$\sum_{n ⩾ 0} n^{2} x^{n}$
Commented by maxmathsup by imad last updated on 07/Jun/19
$let p(x) =Σ_(n=0) ^∞ x^n (∣x∣<1) ⇒p(x) =(1/(1−x)) ⇒p^′ (x) =(1/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n =(x/((x−1)^2 )) ⇒ Σ_(n=1) ^∞ n^2 x^(n−1) ={ (x/((x−1)^2 ))}^((1)) =(((x−1)^2 −x(2(x−1)))/((x−1)^4 )) =((x^2 −2x+1−2x^2 +2x)/((x−1)^4 )) =((1−x^2 )/((1−x)^4 )) =(((1−x)(1+x))/((1−x)^4 )) =((x+1)/((1−x)^3 )) ⇒Σ_(n=1) ^∞ n^2 x^n =((x^2 +x)/((1−x)^3 )) ⇒ Σ_(n≥0) n^2 x^n = ((x^2 +x)/((1−x)^3 )) .$
$l e t p (x) = \sum_{n = 0}^{\infty} x^{n} (∣ x ∣< 1) \Rightarrow p (x) = \frac{1}{1 - x} \Rightarrow p^{^{'}} (x) = \frac{1}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {n x}^{n} = \frac{x}{{(x - 1)}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} x^{n - 1} = {\frac{x}{{(x - 1)}^{2}}}^{(1)} = \frac{{(x - 1)}^{2} - x (2 (x - 1))}{{(x - 1)}^{4}} = \frac{x^{2} - 2 x + 1 - 2 x^{2} + 2 x}{{(x - 1)}^{4}}$ $= \frac{1 - x^{2}}{{(1 - x)}^{4}} = \frac{(1 - x) (1 + x)}{{(1 - x)}^{4}} = \frac{x + 1}{{(1 - x)}^{3}} \Rightarrow \sum_{n = 1}^{\infty} n^{2} x^{n} = \frac{x^{2} + x}{{(1 - x)}^{3}} \Rightarrow$ $\sum_{n ⩾ 0} n^{2} x^{n} = \frac{x^{2} + x}{{(1 - x)}^{3}} .$
Commented by maxmathsup by imad last updated on 07/Jun/19

$s i r a l i i s s a m w h e r e a r e y o u f r o m . . . .$
Commented by aliesam last updated on 07/Jun/19

$i r a q s i r$
Commented by aliesam last updated on 07/Jun/19

$b u t w h y$
Commented by maxmathsup by imad last updated on 07/Jun/19

$f o r n o t h i n g i t s o n l y a i n f o r m a t i o n s i r . . . h o u a r e a l w a y s w e l c o m e . . .$
Commented by aliesam last updated on 07/Jun/19

$o k s i r a n d t h a n k y o u a b o u t t h e g r a e t s o l$
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