For any natural number n1 let an be the largest integer not exceeding √n. Then the value of a₁ + a₂ + …+ a₅₀ is

The correct option is 217.

Let's analyze the problem step-by-step to understand why this is the correct answer.

For any natural number $n$, we are given that $a_n$ is the largest integer not exceeding $\sqrt{n}$. In mathematical notation, this is the floor function of $\sqrt{n}$:
$a_n=\lfloor \sqrt{n}\rfloor$

We want to find the sum:
$S=a_1+a_2+\cdots +a_{50}$

To evaluate this sum, we can group the terms based on the value of $a_n=k$, where $k$ is a positive integer.
The condition $\lfloor \sqrt{n}\rfloor =k$ is equivalent to:
$k\le \sqrt{n}<k+1$

Squaring all parts of the inequality gives:
$k^2\le n<{(k+1)}^2$

Since $n$ and $k$ are integers, this means:
$k^2\le n\le k^2+2k$

The number of such natural numbers $n$ for a given value of $k$ is:
$(k^2+2k)-k^2+1=2k+1$

Let's list the groups of $n$ from $n=1$ up to $n=50$:
• For $k=1$: The values of $n$ are from $1^2=1$ to $2^2-1=3$. There are $2\left(1\right)+1=3$ values: $1,2,3$. For each of these, $a_n=1$.
• For $k=2$: The values of $n$ are from $2^2=4$ to $3^2-1=8$. There are $2\left(2\right)+1=5$ values: $4,5,6,7,8$. For each of these, $a_n=2$.
• For $k=3$: The values of $n$ are from $3^2=9$ to $4^2-1=15$. There are $2\left(3\right)+1=7$ values: $9$ to $15$. For each of these, $a_n=3$.
• For $k=4$: The values of $n$ are from $4^2=16$ to $5^2-1=24$. There are $2\left(4\right)+1=9$ values: $16$ to $24$. For each of these, $a_n=4$.
• For $k=5$: The values of $n$ are from $5^2=25$ to $6^2-1=35$. There are $2\left(5\right)+1=11$ values: $25$ to $35$. For each of these, $a_n=5$.
• For $k=6$: The values of $n$ are from $6^2=36$ to $7^2-1=48$. There are $2\left(6\right)+1=13$ values: $36$ to $48$. For each of these, $a_n=6$.
• For $k=7$: The values of $n$ start at $7^2=49$. Since we only sum up to $n=50$, the only values of $n$ in this group are $49$ and $50$. There are $2$ values. For each of these, $a_n=7$.

Now we calculate the total sum by multiplying each value $k$ by the number of times it appears:
$S=(1\times 3)+(2\times 5)+(3\times 7)+(4\times 9)+(5\times 11)+(6\times 13)+(7\times 2)$

Computing each term:
• $1\times 3=3$
• $2\times 5=10$
• $3\times 7=21$
• $4\times 9=36$
• $5\times 11=55$
• $6\times 13=78$
• $7\times 2=14$

Adding these values together:
$S=3+10+21+36+55+78+14$
$S=13+21+36+55+78+14$
$S=34+36+55+78+14$
$S=70+55+78+14$
$S=125+78+14$
$S=203+14=217$

Thus, the sum is indeed 217.