Problem 161 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The sum of first 9 terms of the series \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots \ldots\) is: \(\quad\) [2015] (A) 96 (B) 142 (C) 192 (D) 71

Step-by-Step Solution

Answer

The sum of the first 9 terms is 96. (Option A)

Step 1: Understanding the Problem

We need to find the sum of the first 9 terms of a series where each term has the form \(\frac{1^3 + 2^3 + \cdots + n^3}{1 + 3 + 5 + \cdots + (2n-1)}\). To solve this, we will first need to understand the pattern or formula for both the numerator and the denominator.

Step 2: Finding Formula for Numerator

The numerator of the general term is the sum of the cubes: \(1^3 + 2^3 + 3^3 + \cdots + n^3\). The formula for the sum of the first \(n\) cubes is \(\left(\frac{n(n+1)}{2}\right)^2\).

Step 3: Finding Formula for Denominator

The denominator is the sum of the first \(n\) odd numbers: \(1 + 3 + 5 + \cdots + (2n-1)\). The sum of the first \(n\) odd numbers is \(n^2\).

Step 4: Simplifying Each Term

Substituting the formulas into the fraction, we have: \[\frac{\left(\frac{n(n+1)}{2}\right)^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}.\]

Step 5: Calculating Each Term's Value

So, each term is simply the square of the next integer divided by 4. Let's calculate this for the first 9 terms: \(\frac{4}{4}, \frac{9}{4}, \frac{16}{4}, \frac{25}{4}, \frac{36}{4}, \frac{49}{4}, \frac{64}{4}, \frac{81}{4}, \frac{100}{4}\).

Step 6: Summing the First 9 Terms

Now add these fractions: \[1 + \frac{9}{4} + 4 + \frac{25}{4} + 9 + \frac{49}{4} + 16 + \frac{81}{4} + 25 = 1 + 2.25 + 4 + 6.25 + 9 + 12.25 + 16 + 20.25 + 25.\] Calculate this to get the total sum.

Step 7: Final Calculation

The overall sum is \(1 + 2.25 + 4 + 6.25 + 9 + 12.25 + 16 + 20.25 + 25 = 96\).

Key Concepts

Sum of Cubes

The sum of cubes is a fascinating topic in mathematics where we focus on the sum of the cubes of the integer numbers. If you have a series of numbers, such as 1, 2, 3, and so on, you can cube each number to form a new series: 1³, 2³, 3³, etc. The sum of these cubed numbers has a specific formula: \(1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2\). This formula gives you the sum when you add up the cubes of the first n natural numbers easily by just inserting n into the formula.

• For example, if n = 3, the cubes are 1³ + 2³ + 3³ = 1 + 8 + 27.
• Using the formula yields: \( \left(\frac{3(3+1)}{2}\right)^2 = \left(\frac{3 \times 4}{2}\right)^2 = 6^2 = 36\); and indeed, 1 + 8 + 27 = 36.

This formula serves as a shortcut to calculate the sum of cubes for any number of terms, making it incredibly useful in sequences like the one in the original exercise.

Odd Numbers Series

Understanding the series of odd numbers is another essential aspect of algebra. Odd numbers are integers not divisible by 2, such as 1, 3, 5, and so forth. A remarkable property of the odd numbers series is that the sum of the first n odd numbers is always equal to \(n^2\). In other words, if you add the first five odd numbers: 1 + 3 + 5 + 7 + 9, you'll get 25, which is 5².

• This characteristic makes it easy to identify the sum of any set number of consecutive odd numbers without direct calculation.
• For instance, for the first 4 odd numbers (1, 3, 5, 7), the sum is \(4^2 = 16\).

In the context of the original problem, this helps simplify the denominator of the fraction, as it quickly tells us the sum for the sequence without much arithmetic.

Problem Solving in Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols; it is a unifying thread of almost all mathematics. When solving problems in algebra, like the one in this exercise, it's important to break down the problem into simpler components. To tackle the original problem: 1. **Identify Patterns**: The problem asks us to recognize patterns in both the numerator (the sum of cubes) and the denominator (sum of odd numbers).

• Recognizing these patterns helps use specific formulas to simplify each term.

2. **Use Algebraic Formulas**: We directly use algebraic formulas, like those for the sum of cubes and the sum of odd numbers, to simplify the expressions. 3. **Simplify and Substitute**: Once the expressions are simplified, you substitute these into your term formula to evaluate specific values. 4. **Perform Arithmetic**: Finally, sum the simplified values step-by-step to obtain the total. By following these structured algebraic methods, we can break down complex-looking problems into simpler parts, allowing for simpler and more efficient problem solving.