### Assume that the symbol $\ddagger x \ddagger$ denotes the largest integer not exceeding $x$. For example, $\ddagger 3 \ddagger = 3$, and $\ddagger 4.9 \ddagger = 4$. What is the value of $\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger.$

$38$

Step by Step Explanation:
1. Given, $\ddagger x \ddagger$ is the largest integer not exceeding $x$.
Let $x$ be a number which lies between two square numbers $a$ and $b$, i.e. $a < x < b$ then $\sqrt{x}$ will lie between $\sqrt{a}$ and $\sqrt{b}$, i.e. $\sqrt{a} < \sqrt{x} < \sqrt{b}$
2. Here, square numbers from $1$ to $16$ are $1, 4, 9, \text{ and } 16$.
Because, the root of the numbers that are greater than or equal to 1 and less than $4, i.e. 1 \le x < 4$ will be greater than or equal to 1 and less than $2, i.e. 1 \le \sqrt{x} < 2$
Therefore, $\ddagger \sqrt{x} \ddagger = 1$ for $1 \le x < 4$
$\implies$$\ddagger \sqrt{ 1 } \ddagger$ $+$ $\ddagger \sqrt{ 2 } \ddagger$ $+$ $\ddagger \sqrt{ 3 } \ddagger$ $=$ $1$ $+$ $1$ $+$ $1$ $= 3 \times 1 = 3$
3. The root of the numbers that are greater than or equal to 4 and less than $9, i.e. 2 \le x < 9$ will be greater than or equal to 2 and less than $3, i.e. 2 \le \sqrt{x} < 3$
Therefore, $\ddagger \sqrt{x} \ddagger = 2$ for $4 \le x < 9$
$\implies$$\ddagger \sqrt{ 4 } \ddagger$ $+$ $\ddagger \sqrt{ 5 } \ddagger$ $+$ $\ddagger \sqrt{ 6 } \ddagger$ $+$ $\ddagger \sqrt{ 7 } \ddagger$ $+$ $\ddagger \sqrt{ 8 } \ddagger$ $=$ $2$ $+$ $2$ $+$ $2$ $+$ $2$ $+$ $2$ $= 5 \times 2 = 10$
4. Similarly, for $9 \le x < 16, \sqrt{x}$ will be $3 \le \sqrt{x} < 4$
Therefore, $\ddagger \sqrt{x} \ddagger = 3$ for $9 \le x < 16$
$\implies$$\ddagger \sqrt{ 9 } \ddagger$ $+$ $\ddagger \sqrt{ 10 } \ddagger$ $+$ $\ddagger \sqrt{ 11 } \ddagger$ $+$ $\ddagger \sqrt{ 12 } \ddagger$ $+$ $\ddagger \sqrt{ 13 } \ddagger$ $+$ $\ddagger \sqrt{ 14 } \ddagger$ $+$ $\ddagger \sqrt{ 15 } \ddagger$ $=$ $3$ $+$ $3$ $+$ $3$ $+$ $3$ $+$ $3$ $+$ $3$ $+$ $3$ $= 7 \times 3 = 21$
And $\sqrt{ 16 } = 4$
$\implies\ddagger\sqrt{ 16 }\ddagger = 4$
5. $\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger = 3 + 10 + 21 + 4$
Hence, the value of $\ddagger \sqrt{1}\ddagger + \ddagger\sqrt{2}\ddagger + \ddagger\sqrt{3}\ddagger \space + ...... + \ddagger\sqrt { 16 }\ddagger$ is $38$