Sum of fraction series
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I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.
Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$
fractions
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
$endgroup$
add a comment |
$begingroup$
I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.
Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$
fractions
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
$endgroup$
$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
4
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23
add a comment |
$begingroup$
I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.
Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$
fractions
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
$endgroup$
I have been trying to solve this question but I am not sure how to, and would appreciate some help. Thanks.
Simplify:
$$frac{1}{4} +frac{1}{4^2} +frac{4^2}{4^3} +frac{4^2}{4^4} +frac{4^4}{4^5} +frac{4^4}{4^6} +ldots +frac{4^{102}}{4^{103}} +frac{4^{102}}{4^{104}}.$$
fractions
fractions
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
edited Dec 8 '18 at 14:19
Rócherz
2,7762721
2,7762721
asked Dec 8 '18 at 14:16
UnknownUnknown
73
asked Dec 8 '18 at 14:16
UnknownUnknown
73
asked Dec 8 '18 at 14:16
UnknownUnknown
73
73
$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
4
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23
add a comment |
$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
4
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23
$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
4
4
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
$endgroup$
add a comment |
$begingroup$
The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
$endgroup$
add a comment |
$begingroup$
The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
$endgroup$
The required sum is $$sum_{r=1}^{52}Bigg(Big(frac{4^{2r-2}}{4^{2r-1}}Big)+Big(frac{4^{2r-2}}{4^{2r}}Big)Bigg)$$
$$=sum_{r=1}^{52}Bigg(Big(frac{1}{4}Big)+Big(frac{1}{4^2}Big)Bigg)$$
$$=52Bigg(frac{1}{4}+frac{1}{4^2}Bigg)$$
$$=13+frac{13}{4}$$
$$=frac{65}{4}$$
Hope it is helpful:)
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
answered Dec 8 '18 at 14:53
MartundMartund
1,593212
1,593212
add a comment |
add a comment |
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$begingroup$
Jimmy: what are the first two terms, and the next two and so on? Can you sum a geometric series?
$endgroup$
– J.G.
Dec 8 '18 at 14:21
4
$begingroup$
You're just repeatedly adding $frac{1}{4}$ and $frac{1}{16}$...
$endgroup$
– KM101
Dec 8 '18 at 14:23