Problem 107 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum \(\left.\frac{3 \times 1^{3}}{1^{2}}+\frac{5 \times\left(1^{3}+2^{3}\right)}{1^{2}+2^{2}}+\frac{7 \times\left(1^{3}+2^{3}+3^{3}\right)}{1^{2}+2^{2}+3^{2}}\right)+\ldots \ldots\) upto \(10^{\text {th }}\) term, is : \(\quad\) [April 10, 2019 (I)] (a) 680 (b) 600 (c) 660 (d) 620

Step-by-Step Solution

Answer

The sum of the series to the 10th term is 680.

Step 1: Understand the General Term

The sequence given is a summation of fractions of the form \( \frac{(2n+1) \times \text{sum of cubes }}{\text{sum of squares}} \). We need to expand these terms for \( n = 1, 2, \ldots, 10 \).

Step 2: Calculate the Numerator for Each Term

The numerator of each term is given by \((2n+1)\times (1^3 + 2^3 + \ldots + n^3)\). We can use the formula for the sum of cubes: \( (1^3 + 2^3 + \ldots + n^3) = \left(\frac{n(n+1)}{2}\right)^2 \).

Step 3: Calculate the Denominator for Each Term

The denominator for each term is \(1^2 + 2^2 + \ldots + n^2\). Use the formula for the sum of squares: \( \frac{n(n+1)(2n+1)}{6} \).

Step 4: Simplify Each Term

Substitute the calculated values from steps 2 and 3 into each term's expression and simplify: \( \frac{(2n+1) \times \left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)(2n+1)}{6}} \). Simplifying this gives \( 3 \times \frac{n(n+1)}{2} \).

Step 5: Sum Up All Terms for n = 1 to 10

We now sum \( 3 \times \frac{n(n+1)}{2} \) from \( n=1 \) to \( n=10 \). This is a straightforward arithmetic summation.

Step 6: Final Calculation

Calculate this sum: \( S = 3 \times \left(\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \ldots + \frac{10(10+1)}{2}\right) \). After computing these summations, we find that the sum is 680.

Key Concepts

Sum of Cubes

When dealing with sequences or series that involve cubic numbers, understanding how to efficiently compute the sum of cubes is crucial. The sum of cubes of the first n natural numbers is expressed with a neat formula: \[ (1^3 + 2^3 + \ldots + n^3) = \left(\frac{n(n+1)}{2}\right)^2 \].
This formula is an interesting result because it states that the sum of cubes up to any integer n is actually the square of the sum of the first n natural numbers. Understanding and using this formula can simplify computations substantially, like in our given sequence, where each term requires summing up cubes.
Here's why this formula is easy and handy to use:

• It reduces the computational steps by compacting individual calculations into a single arithmetic operation.
• You simply calculate the sum of the first n integers, square that sum, and you have the sum of the cubes.
• This formula is derived from the properties of numbers and shows the symmetry and neatness of mathematics.

Sum of Squares

The sum of squares is another powerful tool in mathematics, especially in sequences and series. The sum of the squares of the first n natural numbers can be computed using:\[ 1^2 + 2^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \] The formula stands as a direct computation from basic arithmetic and helps in simplifying many expressions and terms, like denominator computations in this exercise.
Let's explore why this is beneficial:

• The formula reduces large computations into manageable ones, making it straightforward to handle even large values of n easily.
• Calculating each square individually is impractical; the closed formula provides an efficient solution.
• It’s important to remember that this formula is derived through summation techniques and polynomial identities.

Using this formula helps you quickly factor the terms, especially in complex expressions like fractions, where sums of squares are involved in the denominator.

Arithmetic Summation

Arithmetic summation is a foundational technique for solving problems involving series and sequences, like the one given in this exercise. An arithmetic series is one in which the difference between consecutive terms is constant.
The sum of the first n terms of an arithmetic series is computed by the formula:\[ S_n = \frac{n}{2} \times (a + l) \] where \(a\) is the first term and \(l\) is the last term. For this problem, each term is derived from multiplying a factor with the arithmetic sum of n numbers.
Here’s why understanding arithmetic summation is important:

• It simplifies problems by providing a method to directly calculate long sequences without summing each term individually.
• Using this technique ensures time efficiency and simplifies verification checks.
• In many problems, recognizing the structure of the series as arithmetic allows for the application of effective problem-solving techniques.

With these concepts and techniques, you can quickly and efficiently solve many problems involving series, saving time and reducing errors.