Problem 60 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(S_{n}(1 \leq n \leq 9)\) denotes the sum of \(n\) terms of series \(1+22+333+\ldots+999999999\), then for \(2 \leq n \leq 9\) (A) \(S_{n}-S_{n-1}=\frac{1}{9}\left(10^{n}-n^{2}+n\right)\) (B) \(S_{n}=\frac{1}{9}\left(10^{n}-n^{2}+2 n-2\right)\) (C) \(9\left(S_{n}-S_{n-1}\right)=n\left(10^{n}-1\right)\) (D) None of these

Step-by-Step Solution

Answer

Option C is the correct choice.

Step 1: Understanding the Series

The series is composed of numbers like 1, 22, 333, ..., up to n digits each term being composed of the same repeating digit. The nth term is composed of the digit n repeated n times, i.e., the term is \( nn\ldots n \) (n times), which can be expressed as \( n \times \frac{10^n - 1}{9} \).

Step 2: Express the Sum of n Terms

We need to express \( S_n \), which is the sum of the first n terms: \( S_n = 1 + 22 + 333 + \ldots + nn\ldots n \). Each term \( k \) can be written as \( k \times \frac{10^k - 1}{9} \). Thus, \( S_n = \sum_{k=1}^n k \times \frac{10^k - 1}{9} \).

Step 3: Calculate the Difference Between Consecutive Sums

The difference \( S_n - S_{n-1} \) isolates the term when the nth term is first introduced: \( nn\ldots n \). The nth term is given by \( n \times \frac{10^n - 1}{9} \). Thus, \( S_n - S_{n-1} = \frac{1}{9}(n(10^n) - n) \).

Step 4: Check Against Options

Substitute \( S_n - S_{n-1} = \frac{1}{9}(n(10^n) - n) \) into the options to find a match. Simplifying this expression gives \( S_n - S_{n-1} = \frac{n}{9}(10^n - 1) \). Checking option (C), we find: \( 9(S_n - S_{n-1}) = n(10^n - 1) \), which matches perfectly.

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which the difference of any two successive members is a constant. In our original exercise, the series isn't an arithmetic progression in the traditional sense. However, it shares similarities with AP in organizing terms systematically.
Each term increases by repeating the digit corresponding to its sequence. For example, the term "22" has two "2s," the term "333" has three "3s," and so forth. This incremental repetition gives it a patterned sequence akin to AP, yet distinct because each term differs in more than a simple numerical increment.
Understanding this fundamental structure can help you identify and analyze similar patterns. It provides a foundation for grasping series problems and devising methods to calculate sums, like those using specific formulas or terms relationships seen in AP.

Mathematical Induction

Mathematical induction is a powerful proof technique in mathematics, used to prove statements about an infinite number of elements in a sequence. It consists of two main steps:

• **Base Case:** Show the statement is true for the initial step, typically when the index is the smallest value.
• **Inductive Step:** Assume the statement is true for some arbitrary index 'n', and then prove it for the next index 'n + 1'.

In our problem, we are primarily concerned with proving relationships and identities for sums of series. Mathematical induction isn't directly applied in the solution provided, but it acts as a backbone for confirming that the behaviours observed in small cases scale up appropriately. It's about building trust in the formula or series behaviour working universally across all valid assumptions.

Summation Notation

Summation notation is a convenient way to express the sum of terms in a sequence. It uses the sigma (\( \Sigma \)) symbol to signify the summation, followed by an expression for the terms, and specifies the range of indices.
For example, in our exercise, we express the sum of n terms, denoted as \( S_n \), using:\[ S_n = \sum_{k=1}^n k \times \frac{10^k - 1}{9} \]
This concise expression represents a series where each term involves multiplying the sequence number by a fraction ensuring equal repetition pattern.
Understanding how to write and manipulate this notation can greatly simplify solving series-related problems, ensuring clarity and efficiency in finding solutions. It visually condenses complex calculations into an organized statement, helping ensure calculations focus on what matters in reaching the solution.