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Question Number 187546 by yaslm last updated on 18/Feb/23

	Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

	$S = \frac{1 \times 2}{3} + \frac{2 \times 3}{3^{2}} + \frac{3 \times 4}{3^{3}} + \frac{4 \times 5}{3^{4}} + \cdot \cdot \cdot (i)$ $3 S = 1 \times 2 + \frac{2 \times 3}{3} + \frac{3 \times 4}{3^{2}} + \frac{4 \times 5}{3^{3}} + \cdot \cdot \cdot (i i)$ $(i i) - (i)$ $2 S = 2 + \frac{2 \times 2}{3} + \frac{2 \times 3}{3^{2}} + \frac{2 \times 4}{3^{3}} + \frac{2 \times 5}{3^{4}} + \cdot \cdot \cdot (i i i)$ $6 S = 2 \times 3 + 2 \times 2 + \frac{2 \times 3}{3} + \frac{2 \times 4}{3^{2}} + \frac{2 \times 5}{3^{3}} + \cdot \cdot \cdot (i v)$ $(i v) - (i i i)$ $4 S = 8 + \frac{2}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \cdot \cdot \cdot (v)$ $12 S = 24 + 2 + \frac{2}{3} + \frac{2}{3^{2}} + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \cdot \cdot \cdot (v i)$ $12 S - 4 S = 18 \Rightarrow 8 S = 18 \Rightarrow S = \frac{9}{4}$
	Answered by JDamian last updated on 18/Feb/23
	$S = Σ_(k=1) ^(∞) ((k(k+1))/3^k ) f ≡ f(x) = 1+x+x^2 +x^3 + ∙∙∙ = (1/(1−x)) ∀∣x∣<1 D ≡ (d/dx) Df = 1+2x+3x^2 + ∙∙∙ = (1/((1−x)^2 )) ∀∣x∣<1 x^2 Df = 1∙x^2 +2x^3 + ∙∙∙ = (x^2 /((1−x)^2 )) ∀∣x∣<1 g(x)≡D{x^2 Df} = 1∙2x+2∙3x^2 +3∙4x^3 +∙∙∙= = ((2x)/((1−x)^3 )) ∀∣x∣<1 S=g(x)∣_(x=(1/3)) =g((1/3))= =((2×(1/3))/((1−(1/3))^3 ))=(2/(3×((2/3))^3 ))=(2/(3×((2×2^2 )/(3×3^2 )) ))=(9/4)■$
	$S = \underset{k = 1}{\overset{\infty}{Σ}} \frac{k (k + 1)}{3^{k}}$ $f \equiv f (x) = 1 + x + x^{2} + x^{3} + \cdot \cdot \cdot = \frac{1}{1 - x} \forall ∣ x ∣< 1$ $D \equiv \frac{d}{d x}$ $D f = 1 + 2 x + 3 x^{2} + \cdot \cdot \cdot = \frac{1}{{(1 - x)}^{2}} \forall ∣ x ∣< 1$ $x^{2} D f = 1 \cdot x^{2} + 2 x^{3} + \cdot \cdot \cdot = \frac{x^{2}}{{(1 - x)}^{2}} \forall ∣ x ∣< 1$ $g (x) \equiv D {x^{2} D f} = 1 \cdot 2 x + 2 \cdot 3 x^{2} + 3 \cdot 4 x^{3} + \cdot \cdot \cdot =$ $= \frac{2 x}{{(1 - x)}^{3}} \forall ∣ x ∣< 1$ $S = g (x) ∣_{x = \frac{1}{3}} = g (\frac{1}{3}) =$ $= \frac{2 \times \frac{1}{3}}{{(1 - \frac{1}{3})}^{3}} = \frac{2}{3 \times {(\frac{2}{3})}^{3}} = \frac{2}{3 \times \frac{2 \times 2^{2}}{3 \times 3^{2}}} = \frac{9}{4} ◼$
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