Question

For any natural Number 'n', let a_n be the largest number not exceeding \(\sqrt{n}\) , then a1 + a2 + a3... +a50 =

Answer 1

Correct Answer: 217

Let $a_n$ be the largest integer not exceeding $\sqrt{n}$.
We want to compute $\sum_{n=1}^{50} a_n$.
We can find the values of $a_n$ for $n=1, \dots, 50$:
$a_1 = 1$
$a_2 = 1$
$a_3 = 1$
$a_4 = 2$
$a_5 = 2$
$a_6 = 2$
$a_7 = 2$
$a_8 = 2$
$a_9 = 3$
...
$a_{49} = 7$
$a_{50} = 7$
We can group the terms as follows:
$a_n = k$ if $k^2 \le n < (k+1)^2$.
The number of times $k$ appears in the sum is $(k+1)^2 - k^2 = 2k+1$.
We want to find the largest integer $k$ such that $k^2 \le 50$.
$7^2 = 49 \le 50$ and $8^2 = 64 > 50$.
Thus, the largest integer $k$ is 7.
The values of $a_n$ range from 1 to 7.
The number of times $k$ appears in the sum is $(k+1)^2 - k^2 = 2k+1$ for $k=1, \dots, 7$.
The number of times 1 appears is $2(1)+1 = 3$.
The number of times 2 appears is $2(2)+1 = 5$.
The number of times 3 appears is $2(3)+1 = 7$.
The number of times 4 appears is $2(4)+1 = 9$.
The number of times 5 appears is $2(5)+1 = 11$.
The number of times 6 appears is $2(6)+1 = 13$.
The number of times 7 appears is $50 - 49 + 1 = 2$.
The sum is:
$\sum_{n=1}^{50} a_n = 1(3) + 2(5) + 3(7) + 4(9) + 5(11) + 6(13) + 7(2) = 3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$.
Final Answer: The final answer is $\boxed{217}$