∦BeMath∦ (2/5)+(5/(25))+(8/(125))+((11)/(625))+((14)/(3125))+... = ?


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Question Number 107486 by bemath last updated on 11/Aug/20

$∦ B e M a t h ∦$ $\frac{2}{5} + \frac{5}{25} + \frac{8}{125} + \frac{11}{625} + \frac{14}{3125} + . . . = ?$
Commented by bemath last updated on 11/Aug/20

$t h a n k y o u a l l . c o o l l . .$
	Answered by john santu last updated on 11/Aug/20
	$⋇JS⋇ S = (2/5)+(5/(25))+(8/(125))+((11)/(625))+((14)/(3125))+... (1/5)S= (2/(25))+(5/(125))+(8/(625))+((11)/(3125))+... _______________________ − (4/5)S = (2/5)+(3/(25))+(3/(125))+(3/(635))+(3/(3125))+... (4/5)S=(2/5)+3(((1/25)/(1−1/5))) (4/5)S = (2/5)+3((1/(25))×(5/4)) S= (5/4){(2/5)+(3/(20))} = (5/4){((11)/(20))}=((11)/(16))$
	$⋇ JS ⋇$ $S = \frac{2}{5} + \frac{5}{25} + \frac{8}{125} + \frac{11}{625} + \frac{14}{3125} + . . .$ $\frac{1}{5} S = \frac{2}{25} + \frac{5}{125} + \frac{8}{625} + \frac{11}{3125} + . . .$ $_______________________-$ $\frac{4}{5} S = \frac{2}{5} + \frac{3}{25} + \frac{3}{125} + \frac{3}{635} + \frac{3}{3125} + . . .$ $\frac{4}{5} S = \frac{2}{5} + 3 (\frac{1 / 25}{1 - 1 / 5})$ $\frac{4}{5} S = \frac{2}{5} + 3 (\frac{1}{25} \times \frac{5}{4})$ $S = \frac{5}{4} {\frac{2}{5} + \frac{3}{20}} = \frac{5}{4} {\frac{11}{20}} = \frac{11}{16}$
	Answered by Dwaipayan Shikari last updated on 11/Aug/20

	$S = 2 x + 5 x^{2} + 8 x^{3} + 11 x^{4} + . . . . . (x = \frac{1}{5})$ $- x S = - 2 x^{2} - 5 x^{3} - 8 x^{4} - . . .$ $S (1 - x) = 2 x + 3 (x^{2} + x^{3} + x^{4} + . . .)$ $S (1 - x) = 2 x + \frac{3 x^{2}}{1 - x}$ $S = \frac{2 x}{1 - x} + \frac{3 x^{2}}{{(1 - x)}^{2}} = \frac{1}{2} + \frac{3}{16} = \frac{11}{16} (x = \frac{1}{5})$
	Answered by Dwaipayan Shikari last updated on 11/Aug/20

	$\underset{n = 1}{\sum^{\infty}} \frac{3 n - 1}{5^{n}} = 3 \sum^{\infty} \frac{n}{5^{n}} - \sum^{\infty} \frac{1}{5^{n}} = \sum^{\infty} \frac{1}{5^{n}} (\frac{15}{4} - 1) = \frac{1}{4} . \frac{11}{4} = \frac{11}{16}$ $\sum^{\infty} \frac{n}{5^{n}} = \frac{1}{5} + \frac{2}{25} + . . = S (\sum^{\infty} \frac{1}{5^{n}} = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{1}{4})$ $\frac{S}{5} = \frac{1}{25} + \frac{2}{125} + . . .$ $\frac{4 S}{5} = \frac{1}{5} + \frac{1}{25} + . . . = \sum^{\infty} \frac{1}{5^{n}}$ $S = \frac{5}{4} \sum^{\infty} \frac{1}{5^{n}}$
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