Out of control

$f(x) = (x-1)^2+(x-2)^2+(x-3)^2+\cdots+\left(x-10^{10^{100}}+1\right)^2+\left(x-10^{10^{100}}\right)^2$

minima ocuurs at $a$ if $f'(a)=0$ and $f''(a)>0$

$f'(x)=2(x-1)+2(x-2)+2(x-3)+\cdots+2\left(x-10^{10^{100}}+1\right)+2\left(x-10^{10^{100}}\right)$

$\begin{aligned} f''(x)=\underbrace{2+2+2+\cdots+2+2}>0\\{10^{10^{100}} times}\end{aligned}$

now $f'(x)=0$ impiles,

$\begin{aligned}f'(x)&=2(x-1)+2(x-2)+2(x-3)+\cdots+2\left(x-10^{10^{100}}+1\right)+2\left(x-10^{10^{100}}\right)=0\\&\Rightarrow(x-1)+(x-2)+(x-3)+\cdots+\left(x-10^{10^{100}}+1\right)+\left(x-10^{10^{100}}\right)=0\\&\Rightarrow10^{10^{100}}x-\sum_{n=1}^{10^{10^{100}}}n=0\\&\Rightarrow10^{10^{100}}x=\dfrac{10^{10^{100}} \left(10^{10^{100}}+1\right)}{2}\\&\Rightarrow x=\dfrac{ \left(10^{10^{100}}+1\right)}{2}\end{aligned}$

let, $\left(10^{10^{100}}+1\right)=p$ $\hspace{5mm} (p$ is odd)

mimima occurs at $x=\dfrac{p}{2}$

$\begin{aligned}f(\dfrac{p}{2}) &= (\dfrac{p}{2}-1)^2+(\dfrac{p}{2}-2)^2+(\dfrac{p}{2}-3)^2+\cdots+\left(\dfrac{p}{2}-(p-2)\right)^2+\left(\dfrac{p}{2}-(p-1)\right)^2\\&=\dfrac{1}{4} \times\left((p-2)^2+(p-4)^2+(p-6)^2+\cdots+(p-2(p-2))^2+(p-2(p-1))^2\right)\\&=\dfrac{1}{4}\times 2(1^2+3^2+5^2+\cdots+(p-4)^2+(p-2)^2)\\&=\dfrac{1}{2}\times \sum_{n=1}^{\tfrac{p-1}{2}} (2n-1)^2\end{aligned}$

we have,

$\begin{aligned}\sum_{n=1}^x (2n-1)^2&=\dfrac{x(4x^2-1)}{3}\\\Rightarrow f(\dfrac{p}{2})&=\dfrac{1}{2}\times\dfrac{\dfrac{p-1}{2}\left(4\times\left(\dfrac{p-1}{2}\right)^2-1\right)}{3}\\&=\dfrac{1}{12}\times 10^{10^{100}}\times\left(\left(10^{10^{100}}\right)^2-1\right)\\&=\dfrac{1}{12}\times \left(\left(10^{10^{100}}\right)^3-10^{10^{100}}\right)\\&=\dfrac{1}{12}\left(10^{3\cdot10^{100}}-10^{10^{100}}\right)\end{aligned}$

Let's generalize this. We have

$g_n (x) = \displaystyle \sum_{k=1}^n {(x-k)}^2$

Now,

$g'_n (x) = \displaystyle \sum_{k=1}^n 2(x-k) = 0 \implies x = \dfrac{n+1}{2}$

Since $g_n (x)$ has no upper bound, so $x = \dfrac{n+1}{2}$ is the point of minima.

Thus

$\min g_n (x) = \displaystyle \sum_{k=1}^n {\left(\dfrac{n+1}{2}-k \right)}^2 = \dfrac{n (n^2-1)}{12}$

We want to find $f(x) = g_{10^{10^{100}}} (x)$ , so we have

$\min f(x) = \dfrac{10^{10^{100}} \left( 10^{2 \cdot 10^{100}} - 1 \right)}{12} = \dfrac{1}{12}\left(10^{3\cdot10^{100}}-10^{10^{100}}\right)$