Sum of the series $(i)$ and $(ii)$












1














Find the sum of the series:
$(i)1+frac{2}{9}+frac{2.5}{9.18}+frac{2.5.8}{9.18.27}+cdots +infty$
$(ii)1+frac{3}{4}+frac{7}{16}+frac{13}{64}+cdots +infty$
The answer providing my book is :$(i)frac{9}{4}^{frac{1}{3}},(ii)frac{8}{3}$.
i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated.
Thanks in advance.the question






summation summation-method







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  • Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
    – openspace
    Oct 27 '18 at 20:06








  • 1




    It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
    – openspace
    Oct 27 '18 at 20:08










  • the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
    – raihan hossain
    Oct 27 '18 at 20:10










  • @openspace my book didn't provide any information about $a_n$ :)
    – raihan hossain
    Oct 27 '18 at 20:11












  • May I have a look at your book ? It would be nice if you give us good information about task.
    – openspace
    Oct 27 '18 at 20:12
















1














Find the sum of the series:
$(i)1+frac{2}{9}+frac{2.5}{9.18}+frac{2.5.8}{9.18.27}+cdots +infty$
$(ii)1+frac{3}{4}+frac{7}{16}+frac{13}{64}+cdots +infty$
The answer providing my book is :$(i)frac{9}{4}^{frac{1}{3}},(ii)frac{8}{3}$.
i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated.
Thanks in advance.the question






summation summation-method







share|cite|improve this question
























  • Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
    – openspace
    Oct 27 '18 at 20:06








  • 1




    It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
    – openspace
    Oct 27 '18 at 20:08










  • the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
    – raihan hossain
    Oct 27 '18 at 20:10










  • @openspace my book didn't provide any information about $a_n$ :)
    – raihan hossain
    Oct 27 '18 at 20:11












  • May I have a look at your book ? It would be nice if you give us good information about task.
    – openspace
    Oct 27 '18 at 20:12














1












1








1







Find the sum of the series:
$(i)1+frac{2}{9}+frac{2.5}{9.18}+frac{2.5.8}{9.18.27}+cdots +infty$
$(ii)1+frac{3}{4}+frac{7}{16}+frac{13}{64}+cdots +infty$
The answer providing my book is :$(i)frac{9}{4}^{frac{1}{3}},(ii)frac{8}{3}$.
i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated.
Thanks in advance.the question






summation summation-method







share|cite|improve this question















Find the sum of the series:
$(i)1+frac{2}{9}+frac{2.5}{9.18}+frac{2.5.8}{9.18.27}+cdots +infty$
$(ii)1+frac{3}{4}+frac{7}{16}+frac{13}{64}+cdots +infty$
The answer providing my book is :$(i)frac{9}{4}^{frac{1}{3}},(ii)frac{8}{3}$.
i couldn't think of how to start.Because i didn't find any helpful pattern to go further.Any hints or solution will be appreciated.
Thanks in advance.the question






summation summation-method




summation summation-method






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share|cite|improve this question













share|cite|improve this question




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edited Oct 27 '18 at 21:29







raihan hossain

















asked Oct 27 '18 at 19:59









raihan hossainraihan hossain

818




818












  • Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
    – openspace
    Oct 27 '18 at 20:06








  • 1




    It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
    – openspace
    Oct 27 '18 at 20:08










  • the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
    – raihan hossain
    Oct 27 '18 at 20:10










  • @openspace my book didn't provide any information about $a_n$ :)
    – raihan hossain
    Oct 27 '18 at 20:11












  • May I have a look at your book ? It would be nice if you give us good information about task.
    – openspace
    Oct 27 '18 at 20:12


















  • Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
    – openspace
    Oct 27 '18 at 20:06








  • 1




    It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
    – openspace
    Oct 27 '18 at 20:08










  • the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
    – raihan hossain
    Oct 27 '18 at 20:10










  • @openspace my book didn't provide any information about $a_n$ :)
    – raihan hossain
    Oct 27 '18 at 20:11












  • May I have a look at your book ? It would be nice if you give us good information about task.
    – openspace
    Oct 27 '18 at 20:12
















Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
– openspace
Oct 27 '18 at 20:06






Am I right : 1) $sum frac{2 cdot ... cdot (3k+2)}{9 cdot ... cdot 9k}$ ?
– openspace
Oct 27 '18 at 20:06






1




1




It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
– openspace
Oct 27 '18 at 20:08




It's better you to describe the behavior of $a_{n}$. Because it's impossible to determine the series.
– openspace
Oct 27 '18 at 20:08












the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
– raihan hossain
Oct 27 '18 at 20:10




the answer providing my book is $frac{9}{4}^{frac{1}{3}}$. Can you written the sum without summation notation @openspace
– raihan hossain
Oct 27 '18 at 20:10












@openspace my book didn't provide any information about $a_n$ :)
– raihan hossain
Oct 27 '18 at 20:11






@openspace my book didn't provide any information about $a_n$ :)
– raihan hossain
Oct 27 '18 at 20:11














May I have a look at your book ? It would be nice if you give us good information about task.
– openspace
Oct 27 '18 at 20:12




May I have a look at your book ? It would be nice if you give us good information about task.
– openspace
Oct 27 '18 at 20:12










1 Answer
1






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2














I still don't know what it's supposed to be .



1) I supposed that numerator always increased by next even number : $3 = 1 + color{blue}{2}$, then $7 = 3 +color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $displaystyle sum_{k=0}^{infty} frac{k(k+1)+1}{4^{k}} = sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} + sum_{k=0}^{infty} frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $displaystyle S(x) = sum_{k=0}^{infty} frac{x^{k+1}}{4^{k}} = x^{1}sum_{k = 0}^{infty} frac{x^{k}}{4^{k}} = frac{4x^{1}}{4-x}$ , so you have $displaystyle sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (frac{4x}{4-x})''|_{x=1} = frac{32}{27}$






share|cite|improve this answer





















  • According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
    – raihan hossain
    Oct 28 '18 at 9:24










  • @raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
    – openspace
    Oct 28 '18 at 9:35










  • Ok @openspace .Thanks a lot for your hard working sir :)
    – raihan hossain
    Oct 28 '18 at 9:37










  • @raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
    – openspace
    Oct 28 '18 at 9:46










  • Ok @openspace I will wait your answer :)
    – raihan hossain
    Oct 28 '18 at 10:18











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1 Answer
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1 Answer
1






active

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2














I still don't know what it's supposed to be .



1) I supposed that numerator always increased by next even number : $3 = 1 + color{blue}{2}$, then $7 = 3 +color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $displaystyle sum_{k=0}^{infty} frac{k(k+1)+1}{4^{k}} = sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} + sum_{k=0}^{infty} frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $displaystyle S(x) = sum_{k=0}^{infty} frac{x^{k+1}}{4^{k}} = x^{1}sum_{k = 0}^{infty} frac{x^{k}}{4^{k}} = frac{4x^{1}}{4-x}$ , so you have $displaystyle sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (frac{4x}{4-x})''|_{x=1} = frac{32}{27}$






share|cite|improve this answer





















  • According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
    – raihan hossain
    Oct 28 '18 at 9:24










  • @raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
    – openspace
    Oct 28 '18 at 9:35










  • Ok @openspace .Thanks a lot for your hard working sir :)
    – raihan hossain
    Oct 28 '18 at 9:37










  • @raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
    – openspace
    Oct 28 '18 at 9:46










  • Ok @openspace I will wait your answer :)
    – raihan hossain
    Oct 28 '18 at 10:18
















2














I still don't know what it's supposed to be .



1) I supposed that numerator always increased by next even number : $3 = 1 + color{blue}{2}$, then $7 = 3 +color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $displaystyle sum_{k=0}^{infty} frac{k(k+1)+1}{4^{k}} = sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} + sum_{k=0}^{infty} frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $displaystyle S(x) = sum_{k=0}^{infty} frac{x^{k+1}}{4^{k}} = x^{1}sum_{k = 0}^{infty} frac{x^{k}}{4^{k}} = frac{4x^{1}}{4-x}$ , so you have $displaystyle sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (frac{4x}{4-x})''|_{x=1} = frac{32}{27}$






share|cite|improve this answer





















  • According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
    – raihan hossain
    Oct 28 '18 at 9:24










  • @raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
    – openspace
    Oct 28 '18 at 9:35










  • Ok @openspace .Thanks a lot for your hard working sir :)
    – raihan hossain
    Oct 28 '18 at 9:37










  • @raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
    – openspace
    Oct 28 '18 at 9:46










  • Ok @openspace I will wait your answer :)
    – raihan hossain
    Oct 28 '18 at 10:18














2












2








2






I still don't know what it's supposed to be .



1) I supposed that numerator always increased by next even number : $3 = 1 + color{blue}{2}$, then $7 = 3 +color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $displaystyle sum_{k=0}^{infty} frac{k(k+1)+1}{4^{k}} = sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} + sum_{k=0}^{infty} frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $displaystyle S(x) = sum_{k=0}^{infty} frac{x^{k+1}}{4^{k}} = x^{1}sum_{k = 0}^{infty} frac{x^{k}}{4^{k}} = frac{4x^{1}}{4-x}$ , so you have $displaystyle sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (frac{4x}{4-x})''|_{x=1} = frac{32}{27}$






share|cite|improve this answer












I still don't know what it's supposed to be .



1) I supposed that numerator always increased by next even number : $3 = 1 + color{blue}{2}$, then $7 = 3 +color{blue}{4}$ and so on.
And denominator is $4^{k}$.
So we have $displaystyle sum_{k=0}^{infty} frac{k(k+1)+1}{4^{k}} = sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} + sum_{k=0}^{infty} frac{1}{4^{k}}$. The second term is geometric progression and the first one is computed by : $displaystyle S(x) = sum_{k=0}^{infty} frac{x^{k+1}}{4^{k}} = x^{1}sum_{k = 0}^{infty} frac{x^{k}}{4^{k}} = frac{4x^{1}}{4-x}$ , so you have $displaystyle sum_{k=0}^{infty} frac{k(k+1)}{4^{k}} = S''(x)|_{x = 1} = (frac{4x}{4-x})''|_{x=1} = frac{32}{27}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 27 '18 at 21:51









openspaceopenspace

3,4352822




3,4352822












  • According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
    – raihan hossain
    Oct 28 '18 at 9:24










  • @raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
    – openspace
    Oct 28 '18 at 9:35










  • Ok @openspace .Thanks a lot for your hard working sir :)
    – raihan hossain
    Oct 28 '18 at 9:37










  • @raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
    – openspace
    Oct 28 '18 at 9:46










  • Ok @openspace I will wait your answer :)
    – raihan hossain
    Oct 28 '18 at 10:18


















  • According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
    – raihan hossain
    Oct 28 '18 at 9:24










  • @raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
    – openspace
    Oct 28 '18 at 9:35










  • Ok @openspace .Thanks a lot for your hard working sir :)
    – raihan hossain
    Oct 28 '18 at 9:37










  • @raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
    – openspace
    Oct 28 '18 at 9:46










  • Ok @openspace I will wait your answer :)
    – raihan hossain
    Oct 28 '18 at 10:18
















According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
– raihan hossain
Oct 28 '18 at 9:24




According to your solution i get $1.S_{infty} =frac{68}{27}$ :) Is the solution provided by the book is wrong ? @openspace
– raihan hossain
Oct 28 '18 at 9:24












@raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
– openspace
Oct 28 '18 at 9:35




@raihanhossain I see only the answer in book. Its impossible to guess sequence by answer
– openspace
Oct 28 '18 at 9:35












Ok @openspace .Thanks a lot for your hard working sir :)
– raihan hossain
Oct 28 '18 at 9:37




Ok @openspace .Thanks a lot for your hard working sir :)
– raihan hossain
Oct 28 '18 at 9:37












@raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
– openspace
Oct 28 '18 at 9:46




@raihanhossain no problem , I'll think about second series , sequence looks easier , but it's hard to evaluate (for me)
– openspace
Oct 28 '18 at 9:46












Ok @openspace I will wait your answer :)
– raihan hossain
Oct 28 '18 at 10:18




Ok @openspace I will wait your answer :)
– raihan hossain
Oct 28 '18 at 10:18


















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