The total number of elements in the first n rows is:
\[ S = 1 + 2 + 3 + \dots + T_n = \frac{n(n+1)}{2}. \]
To find the row containing 5310, solve:
\[ \frac{n(n+1)}{2} = 5310. \]
Start testing values:
\[ n = 100, \quad T_n = \frac{100 \cdot 101}{2} = 5050. \]
\[ n = 101, \quad T_n = \frac{101 \cdot 102}{2} = 5151. \]
\[ n = 102, \quad T_n = \frac{102 \cdot 103}{2} = 5253. \]
\[ n = 103, \quad T_n = \frac{103 \cdot 104}{2} = 5356. \]
Since 5310 lies between 5253 and 5356, it is in the 103rd row.
Final Answer: 103.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32