Key Concepts and Formulas

1. Limit of a Sum as a Definite Integral: For a continuous function $f(x)$ on $[0,1]$, the limit of a Riemann sum can be expressed as a definite integral: $\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) \, dx$ This is a fundamental tool for evaluating limits of sums by converting them into definite integrals.

2. Sum of an Arithmetic Progression (AP): For an arithmetic progression with $k$ terms, where the first term is $A$ and the last term is $L$, the sum $S_k$ is given by: $S_k = \frac{k}{2}(A + L)$ This formula is essential for simplifying the sum in the denominator of the given limit expression.

3. Power Reduction Formula for Sums of Powers: While not directly a formula, it's a common technique to approximate sums of powers for large $n$. The sum $\sum_{r=1}^n r^a$ is asymptotically proportional to $\frac{n^{a+1}}{a+1}$ as $n \to \infty$. More formally, $\lim_{n \to \infty} \frac{\sum_{r=1}^n r^a}{n^{a+1}} = \frac{1}{a+1}$

Step-by-Step Solution

Let the given limit be $L$. We are given: $L = \mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}$

Step 1: Analyze the Numerator

The numerator is $N = 1^a + 2^a + \dots + n^a = \sum_{r=1}^n r^a$. To relate this to a definite integral, we can rewrite it by factoring out the highest power of $n$: $N = \sum_{r=1}^n r^a = \sum_{r=1}^n n^a \left(\frac{r}{n}\right)^a = n^a \sum_{r=1}^n \left(\frac{r}{n}\right)^a$ Now, we can use the limit of a sum as a definite integral formula. Multiply and divide by $n$: $N = n^{a+1} \left(\frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^a\right)$ As $n \to \infty$, the term in the parenthesis approaches $\int_0^1 x^a \, dx$. $\int_0^1 x^a \, dx = \left[\frac{x^{a+1}}{a+1}\right]_0^1 = \frac{1^{a+1}}{a+1} - \frac{0^{a+1}}{a+1} = \frac{1}{a+1}$ Therefore, for large $n$, the numerator $N$ is asymptotically equivalent to $n^{a+1} \cdot \frac{1}{a+1}$.

Step 2: Analyze the Denominator

The denominator is $D = {(n + 1)}^{a - 1}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]$. Let's first simplify the sum inside the bracket: $S = \left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)$. This is an arithmetic progression with:

• Number of terms ($k$): $n$ (from $na+1$ to $na+n$)
• First term ($A$): $na+1$
• Last term ($L$): $na+n$

Using the sum of an AP formula, $S = \frac{k}{2}(A + L)$: $S = \frac{n}{2}((na+1) + (na+n)) = \frac{n}{2}(2na + n + 1)$ Now, substitute this back into the denominator expression: $D = {(n + 1)}^{a - 1} \cdot \frac{n}{2}(2na + n + 1)$ For large $n$, we can approximate the terms:

• $(n+1)^{a-1} \approx n^{a-1}$
• $2na + n + 1 \approx 2na + n = n(2a+1)$

So, for large $n$: $D \approx n^{a-1} \cdot \frac{n}{2} \cdot n(2a+1) = n^{a-1} \cdot \frac{n^2}{2}(2a+1) = \frac{2a+1}{2} n^{a+1}$

Step 3: Evaluate the Limit

Now we can put the simplified numerator and denominator back into the limit expression: $L = \mathop {\lim }\limits_{n \to \infty } \frac{N}{D} = \mathop {\lim }\limits_{n \to \infty } \frac{n^{a+1} \cdot \frac{1}{a+1}}{\frac{2a+1}{2} n^{a+1}}$ The $n^{a+1}$ terms cancel out: $L = \frac{\frac{1}{a+1}}{\frac{2a+1}{2}} = \frac{1}{a+1} \cdot \frac{2}{2a+1} = \frac{2}{(a+1)(2a+1)}$ We are given that this limit is equal to $\frac{1}{60}$: $\frac{2}{(a+1)(2a+1)} = \frac{1}{60}$ Cross-multiplying gives: $120 = (a+1)(2a+1)$ Expand the right side: $120 = 2a^2 + a + 2a + 1$ $120 = 2a^2 + 3a + 1$ Rearrange into a quadratic equation: $2a^2 + 3a + 1 - 120 = 0$ $2a^2 + 3a - 119 = 0$

Step 4: Solve the Quadratic Equation for 'a'

We need to solve the quadratic equation $2a^2 + 3a - 119 = 0$ for $a$. We can use the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=3$, and $c=-119$. $a = \frac{-3 \pm \sqrt{3^2 - 4(2)(-119)}}{2(2)}$ $a = \frac{-3 \pm \sqrt{9 + 8(119)}}{4}$ $a = \frac{-3 \pm \sqrt{9 + 952}}{4}$ $a = \frac{-3 \pm \sqrt{961}}{4}$ The square root of 961 is 31. $a = \frac{-3 \pm 31}{4}$ We have two possible values for $a$:

1. $a = \frac{-3 + 31}{4} = \frac{28}{4} = 7$
2. $a = \frac{-3 - 31}{4} = \frac{-34}{4} = -\frac{17}{2}$

The problem states that $a$ is a positive real number. Therefore, we choose the positive solution. $a = 7$

Common Mistakes & Tips

• Incorrectly applying the sum to integral formula: Ensure the sum is in the form $\frac{1}{n} \sum f(\frac{r}{n})$ before converting to an integral. Often, you need to manipulate the expression by multiplying and dividing by appropriate powers of $n$.
• Approximation errors in the denominator: When simplifying terms like $(n+1)^{a-1}$ or $na+n+1$ for large $n$, be careful to maintain the dominant terms that affect the highest power of $n$.
• Algebraic errors in solving the quadratic: Double-check the expansion and solution of the quadratic equation, as a small mistake can lead to an incorrect value of $a$.

Summary

The problem involves evaluating a limit that can be transformed into a definite integral for the numerator and simplified using the sum of an arithmetic progression for the denominator. By approximating the dominant terms for large $n$, we found the limit to be $\frac{2}{(a+1)(2a+1)}$. Equating this to $\frac{1}{60}$ led to a quadratic equation in $a$, $2a^2 + 3a - 119 = 0$. Solving this equation and considering the condition that $a$ is positive, we found $a=7$.

The final answer is \boxed{7}. This corresponds to option (A).