Mixing
Sum of series $frac85+frac{16}{65}+cdots cdots +cdots +frac{128}{2^{18}+1}$
$begingroup$
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
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
$endgroup$
add a comment |
$begingroup$
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
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
$endgroup$
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
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
edited Jan 29 at 16:18
Martin Sleziak
44.9k10122276
44.9k10122276
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
asked Jan 4 '17 at 9:04
DXTDXT
6,0662732
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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} $
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
$endgroup$
$begingroup$
@DXT The answer is correct, if you are satisfied with it, please close this problem
$endgroup$
– Samar Imam Zaidi
Jan 28 at 2:07
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
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} $
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
$endgroup$
$begingroup$
@DXT The answer is correct, if you are satisfied with it, please close this problem
$endgroup$
– Samar Imam Zaidi
Jan 28 at 2:07
add a comment |
$begingroup$
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} $
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
$endgroup$
$begingroup$
@DXT The answer is correct, if you are satisfied with it, please close this problem
$endgroup$
– Samar Imam Zaidi
Jan 28 at 2:07
add a comment |
$begingroup$
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} $
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
$endgroup$
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} $
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
answered Jan 27 at 19:36
Samar Imam ZaidiSamar Imam Zaidi
1,5671520
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
add a comment |
$begingroup$
@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
add a comment |
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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