MHB Evaluate the sum of a function

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The discussion focuses on evaluating the sum of the function h(x) defined as h(x) = 9^x / (9^x + 3) for x values ranging from 1/401 to 400/401. A key insight is the relationship h(x) + h(1-x) = 1, which simplifies the evaluation of the sum. By recognizing this symmetry, the sum can be calculated efficiently. Participants express appreciation for the problem's solvability through this observation. The conclusion emphasizes the importance of recognizing functional properties in mathematical evaluations.
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Evaluate $h\left( \dfrac{1}{401} \right)+h\left( \dfrac{2}{401} \right)+\cdots+h\left( \dfrac{400}{401} \right)$ if $h(x)=\dfrac{9^x}{9^x+3}$.
 
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anemone said:
Evaluate $h\left( \dfrac{1}{401} \right)+h\left( \dfrac{2}{401} \right)+\cdots+h\left( \dfrac{400}{401} \right)$ if $h(x)=\dfrac{9^x}{9^x+3}$.

Notice that h(x)+h(1-x)=1.

Hence,

$$h\left(\frac{1}{401}\right)+h\left(\frac{400}{401}\right)=1$$
$$h\left(\frac{2}{401}\right)+h\left(\frac{399}{401}\right)=1$$
$$.$$
$$.$$
$$.$$
$$h\left(\frac{200}{401}\right)+h\left(\frac{201}{401}\right)=1$$

So the sum is 200.
 
Pranav said:
Notice that h(x)+h(1-x)=1.

Hence,

$$h\left(\frac{1}{401}\right)+h\left(\frac{400}{401}\right)=1$$
$$h\left(\frac{2}{401}\right)+h\left(\frac{399}{401}\right)=1$$
$$.$$
$$.$$
$$.$$
$$h\left(\frac{200}{401}\right)+h\left(\frac{201}{401}\right)=1$$

So the sum is 200.

Well done and thanks for participating, Pranav! I think this problem is doable only if one recognizes that in this case, $h(x)+h(1-x)=1$!(Happy)
 
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