Problem 267 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(S_{1}, S_{2}, S_{3}, \ldots, S_{n}\) are the sums of infinite geometric series whose first terms are \(1,2,3, \ldots \ldots, n\) and whose common ratios are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n+1}\) respectively, then find the value of \(S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+\cdots \cdots+S_{2 n-1}{\underline{\phantom{xx}}}^{2}\).

Step-by-Step Solution

Answer

The short answer to the given problem is: \[S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+\cdots +S_{2 n-1}{\underline{\phantom{xx}}}^{2} = \frac{(2n)((2n)+1)(2(2n)+1)}{6} - \frac{n(n)(4n-1)}{3}\]

Step 1: Find the sum of each infinite geometric series

To find the sum of each infinite geometric series, we will use the formula for the sum of an infinite geometric series: \[S = \frac{a}{1 - r}\] where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio. We know that for \(i = 1, 2, 3, ..., n\), the first term is \(i\) and the common ratio is \(\frac{1}{i+1}\). Using this, we can write the sum: \[S_i = \frac{i}{1 - \frac{1}{i+1}}\]

Step 2: Simplify the sum of each infinite geometric series

Next, we have to simplify this expression for \(S_i\). We can start by eliminating the fractions in the denominator: \[S_i = \frac{i}{\frac{i+1-1}{i+1}}\] \[S_i = \frac{i(i+1)}{i}\] \[S_i = i+1\] Now we have the sum for each geometric series: \(S_i = i + 1\)

Step 3: Find the square of each sum

Now we want to find the squared values for each \(S_i\): \[(S_i)^{2} = (i+1)^{2}\]

Step 4: Add the squared sums

We are asked to find the sum of the squares of the first \(2n-1\) terms: \[S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+\cdots +S_{2 n-1}{\underline{\phantom{xx}}}^{2}\] Plug in the squared sum formula from Step 3: \[((1+1)^{2}+(2+1)^{2}+(3+1)^{2}+\cdots +(2n-1+1)^{2})\] \[=(2^{2}+3^{2}+4^{2}+\cdots +(2n)^{2})\] This is the sum of the squares of the first \(2n\) even numbers.

Step 5: Use the sum of squares formula

There is a general formula for the sum of squares of the first \(n\) natural numbers: \[\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}\] Now we will use this formula to find the result. Since we are looking for the squares of the first \(2n\) even numbers, we need to find the sum of squares of the first \(2n\) numbers and subtract the sum of squares of the first \(n\) odd numbers. The sum of squares of the first \(2n\) numbers is given by: \[\frac{(2n)((2n)+1)(2(2n)+1)}{6}\] The sum of squares of the first \(n\) odd numbers can be found by replacing \(i\) with \(2i-1\) in the sum of squares formula: \[\sum_{i=1}^{n}(2i-1)^2 = \frac{n(n)(4n-1)}{3}\] Finally, subtract the sum of squares of odd numbers from the sum of squares of the first \(2n\) numbers:

Step 6: Calculate the result

The required result is: \[\frac{(2n)((2n)+1)(2(2n)+1)}{6} - \frac{n(n)(4n-1)}{3}\] After calculating, we get: \[S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+\cdots +S_{2 n-1}{\underline{\phantom{xx}}}^{2} = \frac{(2n)((2n)+1)(2(2n)+1)}{6} - \frac{n(n)(4n-1)}{3}\]

Key Concepts

Sum of Squares

In mathematics, the sum of squares is a crucial concept with various applications in algebra, calculus, and statistics. Knowing how to calculate it can be very helpful.
When we say "sum of squares," we are usually referring to the sum of squared values of a sequence of numbers. The formula to calculate the sum of the squares of the first \(n\) natural numbers is:

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

This formula helps us quickly find the sum of squares without manually adding each squared value. An important application is in finding the difference between the sum of squares for even and odd sequences. For example, to find the sum of squares of the first \(2n\) numbers, one would use the formula above.
It helps when dealing with sequences where distinguishing between even and odd indexed terms is required. In the solution, we subtract the sum of squares of the first \(n\) odd numbers from the sum of the first \(2n\) numbers to find our answer. This relies on understanding how squares contribute differently to the sum based on their sequence.

Geometric Series Formula

Understanding the geometric series formula is essential when dealing with sequences where each term is a consistent factor of the previous one.
The formula to calculate the sum of an infinite geometric series is:

• \( S = \frac{a}{1 - r} \)

where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
This formula applies when \(|r| < 1\), ensuring the series converges to a finite sum. For example, in our exercise, each series had increasing first terms \(1, 2, 3, \ldots, n\), and decreasing common ratios \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\).
By applying the geometric series formula, we calculate the sum for each series separately, simplifying it by eliminating complex fractions until we obtain the desired expression \(S_i = i + 1\). This systematic approach allows us to manage changing series effectively, emphasizing the power of having a solid formula for infinite series solutions.

Common Ratio

The common ratio is a key element in understanding how a geometric series progresses. In a geometric sequence, the common ratio is the factor that each term is multiplied by to get to the next term.
It is represented by \(r\). For example, in our exercise, the common ratios were \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n+1} \).
The correct identification of the common ratio allows for the accurate application of the geometric series formula. It denotes the rate at which the sequence grows or diminishes. In our problem, the common ratios ensure the terms decrease in size, resulting in convergence to a finite sum.

• The common ratio \(r\) must be \(|r| < 1\) for a series to be convergent and sum to a finite value.
• Understanding how \(r\) impacts the series helps determine if and when the series will approach a sum.

Grasping the idea of a common ratio is fundamental in predicting the behavior of geometric sequences, especially infinite ones. This understanding translates directly into effective problem-solving strategies for finding sums in mathematics.