Find Sum of Solutions to Functional Equation

[anemone]

Given that $(x+1)f(x^2+7x+10)+\left(\dfrac{1-x}{4x}\right)f(x^2+11x+24)=\dfrac{100(x^2+4)}{x}$, find $f(2)+f(3)+\cdots+f(400)$.

 

[jvicens]

Hello,

To solve this problem, we can use the substitution $u=x^2+7x+10$ and $v=x^2+11x+24$. This will transform the original equation into:

$(u+1)f(u)+\left(\dfrac{1-\sqrt{u}}{4\sqrt{u}}\right)f(v)=\dfrac{100(u+4)}{\sqrt{u}}$

Now, we can use the fact that $u+1=(x+1)^2$ and $u+4=(x+4)^2$ to simplify the equation to:

$(x+1)^2f(u)+\left(\dfrac{1-\sqrt{u}}{4\sqrt{u}}\right)f(v)=\dfrac{100(x+4)^2}{\sqrt{u}}$

Next, we can substitute back in for $u$ and $v$ to get:

$(x+1)^2f(x^2+7x+10)+\left(\dfrac{1-\sqrt{x^2+7x+10}}{4\sqrt{x^2+7x+10}}\right)f(x^2+11x+24)=\dfrac{100(x+4)^2}{\sqrt{x^2+7x+10}}$

Finally, we can use the fact that $x+1=2,3,\cdots,400$ to find $f(2)+f(3)+\cdots+f(400)$. Plugging in these values for $x$ and solving the resulting equations will give us the desired sum.

I hope this helps! Let me know if you have any further questions.