Problem 108 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum \(1+\frac{1^{3}+2^{3}}{1+2}+\frac{1^{3}+2^{3}+3^{3}}{1+2+3}+\ldots \ldots+\) \(\frac{1^{3}+2^{3}+3^{3}+\ldots+15^{3}}{1+2+3+\ldots+15}-\frac{1}{2}(1+2+3+\ldots+15\) is equal to : [April 10, 2019 (II)] (a) 620 (b) 1240 (c) 1860 (d) 660

Step-by-Step Solution

Answer

Option (c) 1860.

Step 1: Understanding the Series

The given expression is a series of terms which are of the form: \[\frac{1^3+2^3+3^3+\ldots+n^3}{1+2+3+\ldots+n}\] for terms from \(n = 1\) to \(n = 15\). This sequence of terms involves the sum of cubes divided by the sum of natural numbers.

Step 2: Simplifying Each Term of the Series

Each term can be simplified using the identity: \[1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \] and \[1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}\].Thus, each term becomes: \[\frac{(\frac{n(n+1)}{2})^2}{\frac{n(n+1)}{2}} = \frac{n(n+1)^2}{4}\].

Step 3: Evaluating the Expression

The expression requires adding all the terms from \(n = 1\) to \(n = 15\), and then subtracting half of the sum of numbers from 1 to 15. We compute:\[\sum_{n=1}^{15} \frac{n(n+1)}{4} = \frac{1}{4} \sum_{n=1}^{15} n(n+1)^2.\]Then calculate \(\sum_{n=1}^{15} n(n+1)^2\) which involves expanding to simplify as suitable.

Step 4: Computing the Constants

Calculate \(\sum_{n=1}^{15} n^3 \) using the formula for the sum of cubes: \[\left(\sum_{n=1}^{15} n\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = 3600\].Calculate \(\sum_{n=1}^{15} n = 120\).

Step 5: Final Calculation

Finally, consider the entire expression, substituting the values computed:\[\frac{1}{4} \times 3600 - \frac{1}{2} \times 120 = 900 - 60 = 1860\].

Step 6: Answer Selection

The calculated value of the expression is 1860, which corresponds to choice (c) in the provided options.

Key Concepts

Sum of Cubes

The sum of cubes is a fascinating concept in mathematics, where the cube of each term is summed up. This can be expressed as \(1^3 + 2^3 + 3^3 + \ldots + n^3\). When visualized, it might seem formidable, but a beautiful pattern emerges when you delve deeper.Here's a mathematical identity that simplifies it brilliantly: \[1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\]. This shows that the sum of cubes for the first \(n\) natural numbers is essentially the square of the sum of those numbers themselves.

• It gives an intuitive grasp of symmetry in mathematics.
• It also showcases how numbers interrelate beautifully.

Understanding this identity is crucial, as it simplifies complex series calculations through elegant patterns and relationships.

Series Simplification

Series simplification is a mathematical operation where we break down complex sequences into more manageable components, making them easier to analyze and solve. In the exercise, we're dealing with terms of the form:\[\frac{1^3+2^3+3^3+\ldots+n^3}{1+2+3+\ldots+n}\]. The trick is to simplify each term separately. Using mathematical identities, we can express the sum of cubes and the sum of natural numbers with known formulas. This gives each term a simpler form:\[\frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}} = \frac{n(n+1)}{4}\].

• This reduction allows us to see the broader picture without getting lost in complexity.
• Simplifying series is a powerful tool in both elementary and advanced mathematics.

The process invites us to explore efficiency and elegance in mathematical thinking.

Arithmetic Series

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This consistent difference simplifies the computation of sums and other properties.For example, the sum \(1 + 2 + 3 + \ldots + n\) is an arithmetic series. The sum formula is given by:\[1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}\].
This formula captures the essence of arithmetic series by providing a direct method to calculate the total sum. Applications of arithmetic series extend to various problems involving consecutive integers and is widely utilized in multiple areas of mathematics.

• They are easy to work with once the formula is memorized or understood.
• Arithmetic progressions demonstrate orderly growth and are foundational in various algorithm computations.

Understanding arithmetic series provides solid ground for tackling more complex mathematical challenges.

Mathematical Identities

Mathematical identities are equations true for all variable values within certain constraints. These become invaluable tools in simplifying and solving equations or expressions, especially in series and sequences.In our exercise, the identity for the sum of cubes is:\[1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\].This identity drastically reduces the complexity of several terms by unveiling a direct relationship between summations.

• Mathematical identities often save time by reducing repetitive calculations.
• They display intrinsic symmetry and understanding of how mathematical properties interweave.

Knowing and applying these identities provide a deeper comprehension of numbers, formulating a broad base for further mathematical explorations.