Animated Solution for Mathematics - Functions: Let $[\alpha ]$ denote the greatest integer $\le \alpha$. Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots +[\sqrt{120}]$ is equal to

The Greatest Integer Function (GIF), denoted as $[x]$, represents the floor of a value. We are tasked with evaluating the sum:

Calculating these terms individually is inefficient. Instead, we must identify the underlying structure of the function $f(x)=[\sqrt{x}]$.

The function $f(x)=[\sqrt{x}]$ changes its value only at perfect squares. For any integer $k$, the value of $[\sqrt{x}]$ remains constant at $k$ whenever:

Squaring this inequality yields the condition:

This implies that for all integers $x$ in the interval $[k^2,(k+1{)}^2)$, the value of $[\sqrt{x}]$ is exactly $k$.

We determine the number of integers in each interval by calculating $(k+1{)}^2-k^2$. Expanding this expression gives:

Thus, for a fixed $k$, there are $2k+1$ terms in the sum, each contributing $k$ to the total. The contribution of each group is $k(2k+1)$.

Our series extends to $x=120$. Since $10^2=100$ and $11^2=121$, the group for $k=10$ covers the range $[100,121)$. Because $120$ is the final integer in this range, we can sum from $k=1$ to $k=10$:

We split this into two standard summations:

Using the standard formulas ${\sum }_{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}$ and ${\sum }_{k=1}^{n}k=\frac{n(n+1)}{2}$ with $n=10$:

The final value of the sum is 825.