For any natural number $$ n $$, let $$ a_n $$ be the largest integer not exceeding $$ \sqrt{n} $$. Then, the value of $$ a_1 + a_2 + \cdots + a_{50} $$ is:

Question

Accepted Answer

The problem requires calculating the sum of the largest integers not exceeding the square roots of the first 50 natural numbers. Here's the step-by-step solution:

**Understanding the Problem:**

- For each natural number $$n$$, $$a_n$$ is defined as:
     $$
     a_n = \lfloor \sqrt{n} floor
     $$
     where $$\lfloor x floor$$ represents the largest integer not exceeding $$x$$.

**Group Numbers Based on $$\lfloor \sqrt{n} floor$$:**

- The value of $$a_n$$ remains constant for all $$n$$ in the range:
     $$
     k^2 \leq n < (k+1)^2
     $$
     where $$k$$ is the integer part of $$\sqrt{n}$$.

**Identify Groups for $$n = 1$$ to $$50$$:**

- For $$k = 1$$: $$n = 1, 2, 3$$ (3 terms), $$a_n = 1$$.
   - For $$k = 2$$: $$n = 4, 5, 6, 7, 8$$ (5 terms), $$a_n = 2$$.
   - For $$k = 3$$: $$n = 9, 10, ..., 15$$ (7 terms), $$a_n = 3$$.
   - For $$k = 4$$: $$n = 16, ..., 24$$ (9 terms), $$a_n = 4$$.
   - For $$k = 5$$: $$n = 25, ..., 35$$ (11 terms), $$a_n = 5$$.
   - For $$k = 6$$: $$n = 36, ..., 48$$ (13 terms), $$a_n = 6$$.
   - For $$k = 7$$: $$n = 49, 50$$ (2 terms), $$a_n = 7$$.

**Calculate the Sum of All Terms:**

- Sum for each group:
     - From $$k=1$$: $$3 	imes 1 = 3$$
     - From $$k=2$$: $$5 	imes 2 = 10$$
     - From $$k=3$$: $$7 	imes 3 = 21$$
     - From $$k=4$$: $$9 	imes 4 = 36$$
     - From $$k=5$$: $$11 	imes 5 = 55$$
     - From $$k=6$$: $$13 	imes 6 = 78$$
     - From $$k=7$$: $$2 	imes 7 = 14$$

Total sum:
   $$
   a_1 + a_2 + \cdots + a_{50} = 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217
   $$

### Final Answer:

The value of $$a_1 + a_2 + \cdots + a_{50}$$ is **217**.  
Correct option: **217**

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