Problem 134 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)\left(1^{3}+2^{3}+3^{3}+\ldots+n^{3}\right)}{\left(1^{6}+2^{6}+3^{6}+\ldots+n^{6}\right)}\) is equal to (A) \(\frac{7}{12}\) (B) \(\frac{12}{7}\) (C) \(\frac{5}{12}\) (D) None of these

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{7}{12}\). Option A is correct.

Step 1: Recognize the Formula

We begin by recognizing the formula for the sum of series involved. The formula for the sum of squares up to \(n\) is \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\). The formula for the sum of cubes up to \(n\) is \(\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2\). The formula for the sum of sixth powers up to \(n\) is \(\sum_{k=1}^{n} k^6 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{42}\).

Step 2: Substitute the Formulas

Substitute the formulas for each sum into the limit expression:\[L = \lim_{n \to \infty} \frac{\left(\frac{n(n+1)(2n+1)}{6}\right)\left(\left(\frac{n(n+1)}{2}\right)^2\right)}{\frac{n(n+1)(2n+1)(3n^2+3n-1)}{42}}.\]

Step 3: Simplify the Expression

The numerator becomes:\[\frac{n^2(n+1)^2(2n+1)}{12}\]The denominator is:\[\frac{n(n+1)(2n+1)(3n^2+3n-1)}{42}\]Now simplify the expression as:\[L = \lim_{n \to \infty} \frac{42n^2(n+1)^2(2n+1)}{12n(n+1)(2n+1)(3n^2+3n-1)}.\]

Step 4: Cancel and Simplify Further

Cancel common factors in the numerator and the denominator:\[L = \lim_{n \to \infty} \frac{42n(n+1)}{12(3n^2+3n-1)} = \lim_{n \to \infty} \frac{42n(n+1)}{36n^2 + 36n - 12}.\]

Step 5: Evaluate the Limit

Factor out the highest power of \(n\) from the numerator and denominator:\[L = \lim_{n \to \infty}\frac{42 \left( 1 + \frac{1}{n} \right)}{36 + \frac{36}{n} - \frac{12}{n^2}}.\]As \(n \to \infty\), the terms with \(\frac{1}{n}\), \(\frac{1}{n^2}\) go to zero, so:\[L = \frac{42}{36} = \frac{7}{6}.\]

Key Concepts

Sum of Squares Formula

The sum of squares formula is a useful tool in mathematics, especially when dealing with series and limits. When we want to find the sum of squares from 1 to \( n \), we use the formula: \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}. \]
This formula allows us to compute the total of all squared integers up to \( n \) efficiently without having to square each number individually and add them up.
### Practical ApplicationBy substituting values into the formula, you can determine the sum of squares for any number of terms. For instance, if \( n = 3 \):- Plugging into the formula: \[ \frac{3(3+1)(2 \cdot 3+1)}{6} = \frac{3 \times 4 \times 7}{6} = 14. \] ### Why It MattersUnderstanding this gives insight into more complex problems, such as the one in the exercise, where knowing how to manipulate these series formulae is critical for simplification and evaluation of limits.

• Helps in understanding polynomial growth
• Used in statistical formulas and equations to find variances

Sum of Cubes Formula

Finding the sum of cubes for a sequence involves a different formula than the sum of squares. The sum of cubes from 1 to \( n \) is given by:\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2. \]
This formula is essentially the square of the sum of the first \( n \) natural numbers. ### Intuitive Understanding- If you think about summing numbers and then raising them to the third power (cubing them), this formula tells you it's equivalent to squaring the sum of the sequence directly.- For example, for \( n = 3 \): - First calculate the sum of numbers: \( 1 + 2 + 3 = 6 \), - Then square it: \( 6^2 = 36 \).### Importance in SeriesThe sum of cubes formula is vital in solving problems involving polynomial expansions and evaluating limits. It allows you to transform a complex sequence into a more manageable form. Knowing this helps in simplifying expressions as done in step 2 of the solution by substituting known formulae into expressions.

• Relates to finding volume in geometrical problems
• Integral in higher-level calculus involving polynomial terms

Sum of Powers Formula

The sum of powers formula is a generalized way of summing terms that are raised to a certain power. It covers sums like squares, cubes, and even higher powers. Specifically for the sixth power, used in the original exercise, the formula is:\[ \sum_{k=1}^{n} k^6 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{42}. \]
### Understanding Higher Powers- Each higher power sum has its unique formula, but the structure is often similar, involving products and polynomials of \( n \).- For example: - The formula for \( k^6 \) involves not just quadratic terms but cubic terms (i.e., \( 3n^2 + 3n -1 \)), reflecting the increased complexity.### Application in Simplifying LimitsIn the context of the exercise, using the sum of sixth powers was crucial in matching terms with both the numerator and denominator during simplification.

• Simplifies the evaluation of polynomials in calculus
• Useful in approximations and series expansions

Understanding how these formulas are derived and manipulated allows you to deeply understand the behavior of polynomial functions at theoretical and practical levels.