Evaluate the sum of a function

[anemone]

Evaluate $h\left( \dfrac{1}{401} \right)+h\left( \dfrac{2}{401} \right)+\cdots+h\left( \dfrac{400}{401} \right)$ if $h(x)=\dfrac{9^x}{9^x+3}$.

 

[Saitama]

Evaluate $h\left( \dfrac{1}{401} \right)+h\left( \dfrac{2}{401} \right)+\cdots+h\left( \dfrac{400}{401} \right)$ if $h(x)=\dfrac{9^x}{9^x+3}$.

Spoiler

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.

 

[anemone]

Spoiler

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)