Key Concepts and Formulas

• Greatest Integer Function (Floor Function): $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. This means $k \le \alpha < k+1$ for some integer $k$.
• Sum of first $n$ integers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
• Sum of first $n$ squares: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Step-by-Step Solution

Step 1: Understand the problem and define the sum

We need to evaluate the sum $S = [\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + \ldots + [\sqrt{120}]$. This involves finding the integer part of the square root of each number from 1 to 120 and adding them up.

Step 2: Group terms based on the value of $[\sqrt{n}]$

The key idea is to group the terms in the sum such that $[\sqrt{n}]$ has the same integer value. Let $k = [\sqrt{n}]$, where $k$ is an integer. This means $k \le \sqrt{n} < k+1$. Squaring all parts of the inequality gives $k^2 \le n < (k+1)^2$. Thus, for all $n$ in the interval $[k^2, (k+1)^2)$, we have $[\sqrt{n}] = k$.

Step 3: Determine the range of integer values for $k$

Since $n$ ranges from 1 to 120, we need to find the smallest and largest possible values of $k = [\sqrt{n}]$.

• For $n=1$, $[\sqrt{1}] = 1$, so the smallest value of $k$ is 1.
• For $n=120$, we know that $10^2 = 100 \le 120$ and $11^2 = 121 > 120$. Therefore, $10 \le \sqrt{120} < 11$, so $[\sqrt{120}] = 10$. Thus, the largest value of $k$ is 10. Therefore, $k$ ranges from 1 to 10.

Step 4: Calculate the number of terms for each value of $k$

For each integer $k$ from 1 to 9, we need to find the number of integers $n$ such that $k^2 \le n < (k+1)^2$. The number of integers in this range is $(k+1)^2 - k^2 = k^2 + 2k + 1 - k^2 = 2k+1$.

For $k=10$, we need to find the number of integers $n$ such that $10^2 \le n \le 120$. The number of integers in this range is $120 - 100 + 1 = 21$.

Step 5: Calculate the sum for $k=1$ to $k=9$

For each $k$ from 1 to 9, there are $2k+1$ terms where $[\sqrt{n}] = k$. The sum of these terms is $k(2k+1) = 2k^2 + k$. We need to sum this expression from $k=1$ to $k=9$: $\sum_{k=1}^{9} (2k^2 + k) = 2 \sum_{k=1}^{9} k^2 + \sum_{k=1}^{9} k$ Using the formulas for the sum of the first $n$ integers and the sum of the first $n$ squares with $n=9$: $\sum_{k=1}^{9} k = \frac{9(9+1)}{2} = \frac{9 \times 10}{2} = 45$ $\sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \times 9 + 1)}{6} = \frac{9 \times 10 \times 19}{6} = 3 \times 5 \times 19 = 285$ Therefore, the sum for $k=1$ to $k=9$ is: $2 \times 285 + 45 = 570 + 45 = 615$

Step 6: Calculate the sum for $k=10$

For $k=10$, there are 21 terms, and each term is equal to 10. Therefore, the sum for $k=10$ is $10 \times 21 = 210$.

Step 7: Calculate the total sum

The total sum $S$ is the sum of the contributions from $k=1$ to $k=9$ and the contribution from $k=10$: $S = 615 + 210 = 825$

Common Mistakes & Tips

• Incorrect Upper Limit for $k$: Make sure you correctly identify the maximum integer value of $k$ such that $k^2 \le N$. In this case, $10^2 = 100 \le 120$, and $11^2 = 121 > 120$, so the maximum value of $k$ is 10.
• Miscalculating the number of terms for the last group: Always double-check the number of terms in the last group by subtracting the square of the largest k from N and adding 1: $N - k^2 + 1$.
• Arithmetic Errors: Be careful when calculating the sums using the formulas and when adding the individual contributions.

Summary

The sum $[\sqrt{1}]+[\sqrt{2}]+\ldots+[\sqrt{120}]$ is calculated by grouping terms that have the same integer value for their square root. By summing the contributions $k \times (2k+1)$ for $k=1$ to $9$, and then adding the contribution of $k=10$ for the remaining terms up to $n=120$, we find the total sum. The final answer is 825.

Final Answer

The final answer is \boxed{825}.