Σ_(k=1) ^∞ k^2 (0.8)^(k−1)


Question and Answers Forum

All Questions Topic List
Probability and Statistics Questions
Previous in All Question Next in All Question
Previous in Probability and Statistics Next in Probability and Statistics
Question Number 105487 by Ar Brandon last updated on 29/Jul/20

$\underset{k = 1}{\sum^{\infty}} k^{2} {(0.8)}^{k - 1}$
	Answered by mathmax by abdo last updated on 29/Jul/20
	$let find S(x) =Σ_(n=1) ^∞ n^2 x^(n−1) with ∣x∣<1 we have Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^(n−1) =(d/dx){(x/((x−1)^2 ))} =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) =((x+1)/((1−x)^3 )) ⇒S(x) =((x+1)/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n^2 (0,8)^(n−1) =Σ_(n=1) ^∞ n^2 ((4/5))^(n−1) =S((4/5)) =(((4/5)+1)/((1−(4/5))^3 )) =(9/5)×5^3 =25×9 =225$
	$let find S (x) = \sum_{n = 1}^{\infty} n^{2} x^{n - 1} with ∣ x ∣< 1$ $we have \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow$ $\sum_{n = 1}^{\infty} {nx}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{n - 1} = \frac{d}{dx} {\frac{x}{{(x - 1)}^{2}}}$ $= \frac{{(x - 1)}^{2} - 2 (x - 1) x}{{(x - 1)}^{4}} = \frac{x - 1 - 2 x}{{(x - 1)}^{3}} = \frac{- x - 1}{{(x - 1)}^{3}} = \frac{x + 1}{{(1 - x)}^{3}}$ $\Rightarrow S (x) = \frac{x + 1}{{(1 - x)}^{3}} \Rightarrow$ $\sum_{n = 1}^{\infty} n^{2} {(0, 8)}^{n - 1} = \sum_{n = 1}^{\infty} n^{2} {(\frac{4}{5})}^{n - 1} = S (\frac{4}{5}) = \frac{\frac{4}{5} + 1}{{(1 - \frac{4}{5})}^{3}}$ $= \frac{9}{5} \times 5^{3} = 25 \times 9 = 225$
	Commented by Ar Brandon last updated on 29/Jul/20
	Thanks
	Commented by mathmax by abdo last updated on 29/Jul/20

	$you are welcome .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum