Sum of series $frac85+frac{16}{65}+cdots cdots +cdots +frac{128}{2^{18}+1}$












1












$begingroup$


sum of series $displaystyle frac{8}{5}+frac{16}{65}+cdots cdots +frac{128}{2^{18}+1}$



I have calculate $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1}$



could some help me with this, thanks






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    these are very few terms, you can easily sum them using a simple calculator.
    $endgroup$
    – Math-fun
    Jan 4 '17 at 9:09










  • $begingroup$
    Hint: $2^{5+2}=128$.
    $endgroup$
    – Olivier Oloa
    Jan 4 '17 at 9:11












  • $begingroup$
    I think he means $sum_k a_k$
    $endgroup$
    – Alex
    Jan 4 '17 at 9:11










  • $begingroup$
    There you go. Anything else?
    $endgroup$
    – barak manos
    Jan 4 '17 at 9:16
















1












$begingroup$


sum of series $displaystyle frac{8}{5}+frac{16}{65}+cdots cdots +frac{128}{2^{18}+1}$



I have calculate $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1}$



could some help me with this, thanks






sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    these are very few terms, you can easily sum them using a simple calculator.
    $endgroup$
    – Math-fun
    Jan 4 '17 at 9:09










  • $begingroup$
    Hint: $2^{5+2}=128$.
    $endgroup$
    – Olivier Oloa
    Jan 4 '17 at 9:11












  • $begingroup$
    I think he means $sum_k a_k$
    $endgroup$
    – Alex
    Jan 4 '17 at 9:11










  • $begingroup$
    There you go. Anything else?
    $endgroup$
    – barak manos
    Jan 4 '17 at 9:16














1












1








1





$begingroup$


sum of series $displaystyle frac{8}{5}+frac{16}{65}+cdots cdots +frac{128}{2^{18}+1}$



I have calculate $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1}$



could some help me with this, thanks






sequences-and-series







share|cite|improve this question











$endgroup$




sum of series $displaystyle frac{8}{5}+frac{16}{65}+cdots cdots +frac{128}{2^{18}+1}$



I have calculate $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1}$



could some help me with this, thanks






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 16:18









Martin Sleziak

44.9k10122276




44.9k10122276










asked Jan 4 '17 at 9:04









DXTDXT

6,0662732




6,0662732








  • 3




    $begingroup$
    these are very few terms, you can easily sum them using a simple calculator.
    $endgroup$
    – Math-fun
    Jan 4 '17 at 9:09










  • $begingroup$
    Hint: $2^{5+2}=128$.
    $endgroup$
    – Olivier Oloa
    Jan 4 '17 at 9:11












  • $begingroup$
    I think he means $sum_k a_k$
    $endgroup$
    – Alex
    Jan 4 '17 at 9:11










  • $begingroup$
    There you go. Anything else?
    $endgroup$
    – barak manos
    Jan 4 '17 at 9:16














  • 3




    $begingroup$
    these are very few terms, you can easily sum them using a simple calculator.
    $endgroup$
    – Math-fun
    Jan 4 '17 at 9:09










  • $begingroup$
    Hint: $2^{5+2}=128$.
    $endgroup$
    – Olivier Oloa
    Jan 4 '17 at 9:11












  • $begingroup$
    I think he means $sum_k a_k$
    $endgroup$
    – Alex
    Jan 4 '17 at 9:11










  • $begingroup$
    There you go. Anything else?
    $endgroup$
    – barak manos
    Jan 4 '17 at 9:16








3




3




$begingroup$
these are very few terms, you can easily sum them using a simple calculator.
$endgroup$
– Math-fun
Jan 4 '17 at 9:09




$begingroup$
these are very few terms, you can easily sum them using a simple calculator.
$endgroup$
– Math-fun
Jan 4 '17 at 9:09












$begingroup$
Hint: $2^{5+2}=128$.
$endgroup$
– Olivier Oloa
Jan 4 '17 at 9:11






$begingroup$
Hint: $2^{5+2}=128$.
$endgroup$
– Olivier Oloa
Jan 4 '17 at 9:11














$begingroup$
I think he means $sum_k a_k$
$endgroup$
– Alex
Jan 4 '17 at 9:11




$begingroup$
I think he means $sum_k a_k$
$endgroup$
– Alex
Jan 4 '17 at 9:11












$begingroup$
There you go. Anything else?
$endgroup$
– barak manos
Jan 4 '17 at 9:16




$begingroup$
There you go. Anything else?
$endgroup$
– barak manos
Jan 4 '17 at 9:16










1 Answer
1






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3












$begingroup$

We can write $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1} $ as $displaystyle T_{n} = frac{16.2^{n}}{2^{4n}+4} $



$displaystyle T_{n} = frac{4}{2^{2n}-2.2^{n}+2}-frac{4}{2^{2n}+2.2^{n}+2} $



Adding upto n terms we get
$displaystyle S_{n} = 2-frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic



Solving n=5



$displaystyle S_{5} = 2-frac{4}{1090}=frac{1088}{545} $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @DXT The answer is correct, if you are satisfied with it, please close this problem
    $endgroup$
    – Samar Imam Zaidi
    Jan 28 at 2:07











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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3












$begingroup$

We can write $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1} $ as $displaystyle T_{n} = frac{16.2^{n}}{2^{4n}+4} $



$displaystyle T_{n} = frac{4}{2^{2n}-2.2^{n}+2}-frac{4}{2^{2n}+2.2^{n}+2} $



Adding upto n terms we get
$displaystyle S_{n} = 2-frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic



Solving n=5



$displaystyle S_{5} = 2-frac{4}{1090}=frac{1088}{545} $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @DXT The answer is correct, if you are satisfied with it, please close this problem
    $endgroup$
    – Samar Imam Zaidi
    Jan 28 at 2:07
















3












$begingroup$

We can write $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1} $ as $displaystyle T_{n} = frac{16.2^{n}}{2^{4n}+4} $



$displaystyle T_{n} = frac{4}{2^{2n}-2.2^{n}+2}-frac{4}{2^{2n}+2.2^{n}+2} $



Adding upto n terms we get
$displaystyle S_{n} = 2-frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic



Solving n=5



$displaystyle S_{5} = 2-frac{4}{1090}=frac{1088}{545} $






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @DXT The answer is correct, if you are satisfied with it, please close this problem
    $endgroup$
    – Samar Imam Zaidi
    Jan 28 at 2:07














3












3








3





$begingroup$

We can write $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1} $ as $displaystyle T_{n} = frac{16.2^{n}}{2^{4n}+4} $



$displaystyle T_{n} = frac{4}{2^{2n}-2.2^{n}+2}-frac{4}{2^{2n}+2.2^{n}+2} $



Adding upto n terms we get
$displaystyle S_{n} = 2-frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic



Solving n=5



$displaystyle S_{5} = 2-frac{4}{1090}=frac{1088}{545} $






share|cite|improve this answer









$endgroup$



We can write $displaystyle a_{n} = frac{2^{n+2}}{4^{2n-1}+1} $ as $displaystyle T_{n} = frac{16.2^{n}}{2^{4n}+4} $



$displaystyle T_{n} = frac{4}{2^{2n}-2.2^{n}+2}-frac{4}{2^{2n}+2.2^{n}+2} $



Adding upto n terms we get
$displaystyle S_{n} = 2-frac{4}{2^{2n}+2.2^{n}+2} $ as it is telescopic



Solving n=5



$displaystyle S_{5} = 2-frac{4}{1090}=frac{1088}{545} $







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 27 at 19:36









Samar Imam ZaidiSamar Imam Zaidi

1,5671520




1,5671520












  • $begingroup$
    @DXT The answer is correct, if you are satisfied with it, please close this problem
    $endgroup$
    – Samar Imam Zaidi
    Jan 28 at 2:07


















  • $begingroup$
    @DXT The answer is correct, if you are satisfied with it, please close this problem
    $endgroup$
    – Samar Imam Zaidi
    Jan 28 at 2:07
















$begingroup$
@DXT The answer is correct, if you are satisfied with it, please close this problem
$endgroup$
– Samar Imam Zaidi
Jan 28 at 2:07




$begingroup$
@DXT The answer is correct, if you are satisfied with it, please close this problem
$endgroup$
– Samar Imam Zaidi
Jan 28 at 2:07


















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