Key Concepts and Formulas

• Sum of an Arithmetic Progression: The sum of the first $n$ terms of an arithmetic progression is $S_n = \frac{n}{2}(a_1 + a_n)$, or for the sum of the first $n$ natural numbers, $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$.
• Limits of Rational Functions: For a rational function where the degree of the numerator and denominator are equal, the limit as $n \rightarrow \infty$ is the ratio of the leading coefficients.
• Definite Integral as a Limit of a Sum: A limit of a sum of the form $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)$ can be represented as the definite integral $\int_0^1 f(x) dx$.
• Power Rule of Integration: $\int x^k dx = \frac{x^{k+1}}{k+1} + C$.

Step-by-Step Solution

We need to evaluate two statements, (S1) and (S2), to determine their truth value.

Evaluating Statement (S1):

$\lim_{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)$

• Step 1: Simplify the sum within the parentheses. The sum $2+4+6+\ldots+2n$ is an arithmetic progression. We can factor out $2$: $2(1+2+3+\ldots+n)$ Using the formula for the sum of the first $n$ natural numbers, $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$, the sum becomes: $2 \left( \frac{n(n+1)}{2} \right) = n(n+1)$ Reasoning: We recognized the sum as a standard arithmetic series and applied the known formula for simplification.

• Step 2: Substitute the simplified sum back into the limit expression. The limit expression is now: $\lim_{n \rightarrow \infty} \frac{1}{n^{2}} (n(n+1)) = \lim_{n \rightarrow \infty} \frac{n(n+1)}{n^{2}}$ Reasoning: We replaced the series with its simplified algebraic form to prepare for limit evaluation.

• Step 3: Evaluate the limit of the rational function. Expand the numerator and divide by the highest power of $n$ in the denominator ($n^2$): $\lim_{n \rightarrow \infty} \frac{n^2+n}{n^{2}} = \lim_{n \rightarrow \infty} \left(\frac{n^2}{n^2} + \frac{n}{n^2}\right) = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)$ As $n \rightarrow \infty$, $\frac{1}{n} \rightarrow 0$. Therefore, the limit is: $1 + 0 = 1$ Reasoning: We simplified the rational expression and used the property that $\lim_{n \rightarrow \infty} \frac{c}{n^k} = 0$ for $k>0$ to evaluate the limit.

• Conclusion for (S1): The calculated limit is $1$, which matches the value given in statement (S1). Thus, (S1) is true.

Evaluating Statement (S2):

$\lim_{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)$

• Step 1: Rewrite the sum using sigma notation and prepare for definite integral conversion. The sum can be written as $\sum_{r=1}^{n} r^{15}$. The limit becomes: $\lim_{n \rightarrow \infty} \frac{1}{n^{16}} \sum_{r=1}^{n} r^{15}$ To use the definite integral as a limit of a sum formula, we need a $\frac{1}{n}$ outside the summation. We split $n^{16}$ into $n \cdot n^{15}$: $\lim_{n \rightarrow \infty} \frac{1}{n} \cdot \frac{1}{n^{15}} \sum_{r=1}^{n} r^{15}$ Now, we distribute the $\frac{1}{n^{15}}$ into the summation by dividing each term $r^{15}$ by $n^{15}$: $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r^{15}}{n^{15}} = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^{15}$ Reasoning: We algebraically manipulated the expression to match the standard form of a Riemann sum, $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)$.

• Step 2: Convert the limit of the sum to a definite integral. By comparing $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^{15}$ with $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)$, we identify $f(x) = x^{15}$. The limits of integration are from $0$ to $1$, because as $n \rightarrow \infty$, $\frac{r}{n}$ ranges from $\frac{1}{n} \rightarrow 0$ to $\frac{n}{n} = 1$. The definite integral is: $\int_0^1 x^{15} dx$ Reasoning: We applied the definition of the definite integral as the limit of a Riemann sum, identifying the integrand and the integration interval.

• Step 3: Evaluate the definite integral. Using the power rule of integration, $\int x^k dx = \frac{x^{k+1}}{k+1}$: $\int_0^1 x^{15} dx = \left[ \frac{x^{16}}{16} \right]_0^1$ Evaluating at the limits: $\frac{(1)^{16}}{16} - \frac{(0)^{16}}{16} = \frac{1}{16} - 0 = \frac{1}{16}$ Reasoning: We used the Fundamental Theorem of Calculus to evaluate the definite integral.

• Conclusion for (S2): The calculated limit is $\frac{1}{16}$, which matches the value given in statement (S2). Thus, (S2) is true.

Common Mistakes & Tips

• Algebraic Errors in Sum Simplification: Carefully check the formula for the sum of series and the algebraic manipulation, especially when factoring and expanding.
• Incorrect Riemann Sum Conversion: Ensure that exactly one $\frac{1}{n}$ term is outside the summation. If the expression is $\frac{1}{n^k} \sum f(\frac{r}{n})$ with $k \neq 1$, you must carefully split $n^k$ to isolate a single $\frac{1}{n}$.
• Identifying $f(x)$: When converting to an integral, make sure $f(x)$ is correctly identified by replacing $\frac{r}{n}$ with $x$.

Summary

Statement (S1) was evaluated by simplifying the arithmetic progression and then computing the limit of the resulting rational function. This yielded a limit of 1, confirming (S1) is true. Statement (S2) was evaluated by converting the limit of the sum into a definite integral using the Riemann sum definition. Evaluating the integral $\int_0^1 x^{15} dx$ resulted in $\frac{1}{16}$, confirming (S2) is true. Since both statements are true, the correct option is (B).

The final answer is $\boxed{A}$