Problem 109 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The sum of the series \(1+2 \times 3+3 \times 5+4 \times 7+\ldots \ldots\) upto \(11^{\text {th }}\) term is: [April 09, 2019 (II)] (a) 915 (b) 946 (c) 945 (d) 916

Step-by-Step Solution

Answer

The sum of the series up to the 11th term is 946 (option b).

Step 1: Identify the Pattern in the Series

The given series starts with the terms: \(1, 2 \times 3, 3 \times 5, 4 \times 7, \ldots \). Observing these terms, each term \(T_n\) can be written as \(n \times (2n - 1)\). This gives the terms as \(1 \times 1, 2 \times 3, 3 \times 5, 4 \times 7, \ldots\).

Step 2: Express the General Term

Express the general term \(T_n\) of the series as follows: \[ T_n = n \times (2n-1) \].

Step 3: Calculate the Sum of the Series

The sum of the series up to the 11th term is given by \( \sum_{n=1}^{11} T_n \). Therefore, calculate: \( \sum_{n=1}^{11} n \times (2n-1) \).

Step 4: Expand and Simplify the Sum

The expression \( n \times (2n-1) \) can be expanded to \( 2n^2 - n \). Hence, \( \sum_{n=1}^{11} (2n^2 - n) \) becomes \( 2 \sum_{n=1}^{11} n^2 - \sum_{n=1}^{11} n \).

Step 5: Use Formula for Sum of Natural Numbers and Sum of Squares

Use the formula \( \sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} \) and \( \sum_{n=1}^{N} n = \frac{N(N+1)}{2} \) to calculate the values. For \( N = 11 \), this results in: \( \sum_{n=1}^{11} n^2 = \frac{11 \times 12 \times 23}{6} = 506 \) and \( \sum_{n=1}^{11} n = \frac{11 \times 12}{2} = 66 \).

Step 6: Complete the Calculation

Substitute these values back into the equation: \( 2 \times 506 - 66 = 1012 - 66 = 946 \).

Step 7: Identify the Correct Option

The calculated sum of the series up to the 11th term is 946. Therefore, the correct answer is option (b) 946.

Key Concepts

Sum of Natural Numbers

Understanding the sum of natural numbers is crucial as it forms the foundation for many mathematical series.
Natural numbers are the numbers we count naturally: 1, 2, 3, 4, and so on.
To find the sum of the first \( N \) natural numbers, we use the formula:

• \( \sum_{n=1}^{N} n = \frac{N(N+1)}{2} \)

This formula comes from the idea of pairing numbers in the sequence to make uniform sums.
For example, if you list the numbers from 1 to 10, you can pair them as (1+10), (2+9), (3+8), up to (5+6).
Each pair sums to 11, and since there are 5 such pairs, the sum is \( 5 \times 11 = 55 \).
This pattern holds for any \( N \), which is why the formula is reliable and efficient.
Using this formula, you can quickly determine the sum of numbers without adding them individually, saving both time and effort.

Sum of Squares

The sum of squares is slightly more complex than the sum of natural numbers, but it is just as important.
It is often used in statistics, physics, and other areas to measure variance or energy, among other things.
The formula for the sum of squares for the first \( N \) numbers is:

• \( \sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} \)

This formula gives the total when each of the first \( N \) numbers is squared, then added together.
To understand this formula, consider breaking it into parts:

• The term \( N(N+1) \) is somewhat similar to the formula for the sum of natural numbers but adjusted to account for squaring each number.
• The additional factor \( 2N+1 \) helps correct the increasing difference in value as numbers grow larger.

Using this formula, figuring out the sum of squares is greatly simplified and removes the need for manual addition or error-prone arithmetic.

General Term in a Series

Finding the general term of a series helps in understanding the pattern quickly and in computing larger sequences efficiently.
In the original exercise, each term \( T_n \) can be described by a single formula:

• \( T_n = n \times (2n-1) \)

This captures the regular pattern of numbers multiplying within the series.
Recognizing such patterns isn't always straightforward, but once done, it simplifies the summing process significantly.
Knowing the general term is valuable because it allows you to:

• Predict any term in the series without having to write out all previous ones.
• Sum the entire series up to any given term more systematically.

Thus, having a formula to express the general term is a powerful tool that enhances not only your problem-solving skills but also your understanding of sequence behavior in mathematics.