Calculate, $fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots fbigg(frac19961997bigg)$ [duplicate]
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This question already has an answer here:
find $f(frac12014)+f(frac22014)+…+f(frac20132014)$ of $f(x)=frac22+4^x$
1 answer
If $$f(x)=frac4^x4^x+2$$
Calculate,
$$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)$$
My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$fbigg(frac12bigg)=frac12$$
$$fbigg(frac13bigg)+fbigg(frac23bigg)=1$$
$$fbigg(frac14bigg)+fbigg(frac24bigg)+fbigg(frac34bigg)=frac32$$
I could see that,
$$fbigg(frac1nbigg)+fbigg(frac2nbigg)+fbigg(frac3nbigg)ldots
fbigg(fracn-1nbigg)=fracn-12$$
So, $$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?
calculus functions
asked Aug 21 at 13:26
prog_SAHIL
1,197318
marked as duplicate by lab bhattacharjee calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Aug 21 at 14:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
8
down vote
favorite
This question already has an answer here:
find $f(frac12014)+f(frac22014)+…+f(frac20132014)$ of $f(x)=frac22+4^x$
1 answer
If $$f(x)=frac4^x4^x+2$$
Calculate,
$$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)$$
My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$fbigg(frac12bigg)=frac12$$
$$fbigg(frac13bigg)+fbigg(frac23bigg)=1$$
$$fbigg(frac14bigg)+fbigg(frac24bigg)+fbigg(frac34bigg)=frac32$$
I could see that,
$$fbigg(frac1nbigg)+fbigg(frac2nbigg)+fbigg(frac3nbigg)ldots
fbigg(fracn-1nbigg)=fracn-12$$
So, $$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?
calculus functions
asked Aug 21 at 13:26
prog_SAHIL
1,197318
marked as duplicate by lab bhattacharjee calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Aug 21 at 14:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
This question already has an answer here:
find $f(frac12014)+f(frac22014)+…+f(frac20132014)$ of $f(x)=frac22+4^x$
1 answer
If $$f(x)=frac4^x4^x+2$$
Calculate,
$$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)$$
My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$fbigg(frac12bigg)=frac12$$
$$fbigg(frac13bigg)+fbigg(frac23bigg)=1$$
$$fbigg(frac14bigg)+fbigg(frac24bigg)+fbigg(frac34bigg)=frac32$$
I could see that,
$$fbigg(frac1nbigg)+fbigg(frac2nbigg)+fbigg(frac3nbigg)ldots
fbigg(fracn-1nbigg)=fracn-12$$
So, $$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?
calculus functions
asked Aug 21 at 13:26
prog_SAHIL
1,197318
This question already has an answer here:
find $f(frac12014)+f(frac22014)+…+f(frac20132014)$ of $f(x)=frac22+4^x$
1 answer
If $$f(x)=frac4^x4^x+2$$
Calculate,
$$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)$$
My Attempt:
I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,
$$fbigg(frac12bigg)=frac12$$
$$fbigg(frac13bigg)+fbigg(frac23bigg)=1$$
$$fbigg(frac14bigg)+fbigg(frac24bigg)+fbigg(frac34bigg)=frac32$$
I could see that,
$$fbigg(frac1nbigg)+fbigg(frac2nbigg)+fbigg(frac3nbigg)ldots
fbigg(fracn-1nbigg)=fracn-12$$
So, $$fbigg(frac11997bigg)+fbigg(frac21997bigg)+fbigg(frac31997bigg)ldots
fbigg(frac19961997bigg)=998$$
which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?
This question already has an answer here:
find $f(frac12014)+f(frac22014)+…+f(frac20132014)$ of $f(x)=frac22+4^x$
1 answer
calculus functions
asked Aug 21 at 13:26
prog_SAHIL
1,197318
asked Aug 21 at 13:26
prog_SAHIL
1,197318
asked Aug 21 at 13:26
prog_SAHIL
1,197318
asked Aug 21 at 13:26
prog_SAHIL
1,197318
1,197318
marked as duplicate by lab bhattacharjee calculus
Users with the  calculus badge can single-handedly close calculus questions as duplicates and reopen them as needed.
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Aug 21 at 14:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by lab bhattacharjee calculus
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Aug 21 at 14:04
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35
add a comment |Â
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
I would say your method is practically speaking what I would also do.
Maybe I would rephrase it as follows:
Claim: $f(a)+f(1-a)=1$.
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.
answered Aug 21 at 13:33
A. Pongrácz
4,112625
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
 |Â
show 1 more comment
up vote
4
down vote
Given $f(x)=dfrac4^x4^x+2$
From that we get $f(1-x)=dfrac24^x+2$
First let us take the last term $fleft(dfrac11997right)$
Notice that $fleft(dfrac11997right)=fleft(1-dfrac11997right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$fleft(dfrac11997right)+fleft(dfrac21997right)+........+fleft(1-dfrac21997+fleft(1-dfrac11997right)right)$
all of them makes pairs.
So, the total pairs $=dfrac19962=998$
So, the sum $=1+1+1+......998$ times $=998$
answered Aug 21 at 13:37
Key Flex
1
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
I would say your method is practically speaking what I would also do.
Maybe I would rephrase it as follows:
Claim: $f(a)+f(1-a)=1$.
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.
answered Aug 21 at 13:33
A. Pongrácz
4,112625
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
 |Â
show 1 more comment
up vote
7
down vote
accepted
I would say your method is practically speaking what I would also do.
Maybe I would rephrase it as follows:
Claim: $f(a)+f(1-a)=1$.
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.
answered Aug 21 at 13:33
A. Pongrácz
4,112625
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
 |Â
show 1 more comment
up vote
7
down vote
accepted
up vote
7
down vote
accepted
I would say your method is practically speaking what I would also do.
Maybe I would rephrase it as follows:
Claim: $f(a)+f(1-a)=1$.
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.
answered Aug 21 at 13:33
A. Pongrácz
4,112625
I would say your method is practically speaking what I would also do.
Maybe I would rephrase it as follows:
Claim: $f(a)+f(1-a)=1$.
Then write $S$ for the sum in question, and then $2S$ can be written as $f(1/1997+1996/1997) + cdots$ (the Gauss trick), which is $1996$ by the claim, so $S=998$.
answered Aug 21 at 13:33
A. Pongrácz
4,112625
answered Aug 21 at 13:33
A. Pongrácz
4,112625
answered Aug 21 at 13:33
A. Pongrácz
4,112625
answered Aug 21 at 13:33
A. Pongrácz
4,112625
4,112625
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
 |Â
show 1 more comment
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
2
2
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
literally 30 seconds before me sheesh
– Rushabh Mehta
Aug 21 at 13:33
1
1
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
Are you sure of it? I got a different result.
– Dr. Sonnhard Graubner
Aug 21 at 13:34
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
@Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get
– Rushabh Mehta
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation.
– A. Pongrácz
Aug 21 at 13:36
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
Ok, i will also check my result, before posting.
– Dr. Sonnhard Graubner
Aug 21 at 13:38
 |Â
show 1 more comment
up vote
4
down vote
Given $f(x)=dfrac4^x4^x+2$
From that we get $f(1-x)=dfrac24^x+2$
First let us take the last term $fleft(dfrac11997right)$
Notice that $fleft(dfrac11997right)=fleft(1-dfrac11997right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$fleft(dfrac11997right)+fleft(dfrac21997right)+........+fleft(1-dfrac21997+fleft(1-dfrac11997right)right)$
all of them makes pairs.
So, the total pairs $=dfrac19962=998$
So, the sum $=1+1+1+......998$ times $=998$
answered Aug 21 at 13:37
Key Flex
1
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
add a comment |Â
up vote
4
down vote
Given $f(x)=dfrac4^x4^x+2$
From that we get $f(1-x)=dfrac24^x+2$
First let us take the last term $fleft(dfrac11997right)$
Notice that $fleft(dfrac11997right)=fleft(1-dfrac11997right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$fleft(dfrac11997right)+fleft(dfrac21997right)+........+fleft(1-dfrac21997+fleft(1-dfrac11997right)right)$
all of them makes pairs.
So, the total pairs $=dfrac19962=998$
So, the sum $=1+1+1+......998$ times $=998$
answered Aug 21 at 13:37
Key Flex
1
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Given $f(x)=dfrac4^x4^x+2$
From that we get $f(1-x)=dfrac24^x+2$
First let us take the last term $fleft(dfrac11997right)$
Notice that $fleft(dfrac11997right)=fleft(1-dfrac11997right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$fleft(dfrac11997right)+fleft(dfrac21997right)+........+fleft(1-dfrac21997+fleft(1-dfrac11997right)right)$
all of them makes pairs.
So, the total pairs $=dfrac19962=998$
So, the sum $=1+1+1+......998$ times $=998$
answered Aug 21 at 13:37
Key Flex
1
Given $f(x)=dfrac4^x4^x+2$
From that we get $f(1-x)=dfrac24^x+2$
First let us take the last term $fleft(dfrac11997right)$
Notice that $fleft(dfrac11997right)=fleft(1-dfrac11997right)$ and same for the rest of the terms.
Now, $f(x)+f(1-x)=1$
$fleft(dfrac11997right)+fleft(dfrac21997right)+........+fleft(1-dfrac21997+fleft(1-dfrac11997right)right)$
all of them makes pairs.
So, the total pairs $=dfrac19962=998$
So, the sum $=1+1+1+......998$ times $=998$
answered Aug 21 at 13:37
Key Flex
1
answered Aug 21 at 13:37
Key Flex
1
answered Aug 21 at 13:37
Key Flex
1
answered Aug 21 at 13:37
Key Flex
1
1
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
add a comment |Â
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
The reason I suggested to write down $S$ twice instead of pairing, is because that way you do not have to discuss anything about the case of an odd number of terms. But in any case, this is practically the same answer I gave.
– A. Pongrácz
Aug 21 at 13:40
add a comment |Â
I would write $$f(x)=frac2^2x2+2^2x$$
– Dr. Sonnhard Graubner
Aug 21 at 13:31
I tried simplifying as $$1-frac22^2x+2$$ but no help.
– prog_SAHIL
Aug 21 at 13:32
And this is $$1-frac12^2x-1+1$$
– Dr. Sonnhard Graubner
Aug 21 at 13:35