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If $$f(x)=\frac{4^x}{4^x+2}$$

Calculate,

$$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)$$

My Attempt:

I was not able to generalise the expression or get a solid pattern, so I started with smaller numbers and calculated,

$$f\bigg(\frac{1}{2}\bigg)=\frac{1}{2}$$

$$f\bigg(\frac{1}{3}\bigg)+f\bigg(\frac{2}{3}\bigg)=1$$

$$f\bigg(\frac{1}{4}\bigg)+f\bigg(\frac{2}{4}\bigg)+f\bigg(\frac{3}{4}\bigg)=\frac{3}{2}$$

I could see that,

$$f\bigg(\frac{1}{n}\bigg)+f\bigg(\frac{2}{n}\bigg)+f\bigg(\frac{3}{n}\bigg)\ldots f\bigg(\frac{n-1}{n}\bigg)=\frac{n-1}{2}$$

So, $$f\bigg(\frac{1}{1997}\bigg)+f\bigg(\frac{2}{1997}\bigg)+f\bigg(\frac{3}{1997}\bigg)\ldots f\bigg(\frac{1996}{1997}\bigg)=998$$

which is indeed the right answer. But I am not satisfied with my method. How else can I solve it?

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  • $\begingroup$ I would write $$f(x)=\frac{2^{2x}}{2+2^{2x}}$$ $\endgroup$ Aug 21, 2018 at 13:31
  • $\begingroup$ I tried simplifying as $$1-\frac{2}{2^{2x}+2}$$ but no help. $\endgroup$
    – prog_SAHIL
    Aug 21, 2018 at 13:32
  • $\begingroup$ And this is $$1-\frac{1}{2^{2x-1}+1}$$ $\endgroup$ Aug 21, 2018 at 13:35

2 Answers 2

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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 $S=998$.

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    $\begingroup$ literally 30 seconds before me sheesh $\endgroup$ Aug 21, 2018 at 13:33
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    $\begingroup$ Are you sure of it? I got a different result. $\endgroup$ Aug 21, 2018 at 13:34
  • $\begingroup$ @Dr.SonnhardGraubner I'm pretty sure this is correct...what result did you get $\endgroup$ Aug 21, 2018 at 13:36
  • $\begingroup$ Rushabh Mehta: sorry, bad luck. Dr. Sonnhard Graubner: I checked it twice, but if you want, I can provide the calculation. $\endgroup$ Aug 21, 2018 at 13:36
  • $\begingroup$ Ok, i will also check my result, before posting. $\endgroup$ Aug 21, 2018 at 13:38
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Given $f(x)=\dfrac{4^x}{4^x+2}$

From that we get $f(1-x)=\dfrac{2}{4^x+2}$

First let us take the last term $f\left(\dfrac{1}{1997}\right)$

Notice that $f\left(\dfrac{1}{1997}\right)=f\left(1-\dfrac{1}{1997}\right)$ and same for the rest of the terms.

Now, $f(x)+f(1-x)=1$

$f\left(\dfrac{1}{1997}\right)+f\left(\dfrac{2}{1997}\right)+........+f\left(1-\dfrac{2}{1997}+f\left(1-\dfrac{1}{1997}\right)\right)$

all of them makes pairs.

So, the total pairs $=\dfrac{1996}{2}=998$

So, the sum $=1+1+1+......998$ times $=998$

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  • $\begingroup$ 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. $\endgroup$ Aug 21, 2018 at 13:40

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