Question: How can we confirm that the local extrema we have found by putting the derivative of the function equal to zero are absolute extrema or not ? if they are not absolute extrema then how can we find the absolute extrema ?

Answer: 
Since 
$$(x-i)(x-i-1)=\frac13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)$$
one has
$$\begin{align}\sum\_{i=0}^{k}\binom{x-i}{2}&=\sum\_{i=0}^{k}\frac{(x-i)(x-i-1)}{2}\\&=\frac 12\sum\_{i=0}^{k}\frac 13\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)\\&=\frac 16\sum\_{i=0}^{k}\left((x-i-1)(x-i)(x-i+1)-(x-i-2)(x-i-1)(x-i)\right)\\&=\frac 16\left((x-1)x(x+1)-(x-k-2)(x-k-1)(x-k)\right)\\&=\frac{k+1}{2}x^2-\frac{(k+1)^2}{2}x+\frac{k(k+1)(k+2)}{6}\end{align}$$
$$\frac{1}{2}\sum\_{i=0}^k (x-i)(x-i-1) = \frac{1}{2} \left( \sum\_{i=0}^k (x^2-2xi+i^2 + x-i) \right) =$$
$$=\frac{1}{2} \left( \sum\_{i=0}^k (x^2+ x) + \sum\_{i=0}^k (-2xi-i) + \sum\_{i=0}^k i^2\right)= \frac{1}{2} \left( (k+1)(x^2+x) -(2x+1)\left(\frac{k(k+1)}{2} \right) + \frac{k(k+1)(2k+1)}{6}\right)$$