The sum of the function \( h(x) = \frac{9^x}{9^x + 3} \) evaluated from \( h\left( \frac{1}{401} \right) \) to \( h\left( \frac{400}{401} \right) \) can be simplified using the property \( h(x) + h(1-x) = 1 \). This property allows for pairing terms in the sum, resulting in \( 200 \) pairs, each summing to \( 1 \). Therefore, the total sum is \( 200 \), confirming the effectiveness of recognizing functional symmetries in solving such problems.
PREREQUISITESMathematics students, educators, and anyone interested in functional analysis and problem-solving techniques in algebra.
anemone said:Evaluate $h\left( \dfrac{1}{401} \right)+h\left( \dfrac{2}{401} \right)+\cdots+h\left( \dfrac{400}{401} \right)$ if $h(x)=\dfrac{9^x}{9^x+3}$.
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.