Problem 6 - A Complete Resource Book in Mathematics for JEE Main 2017

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 correct: \(9(S_n - S_{n-1}) = n(10^n - 1)\).

Step 1: Understanding the Problem

We are asked to find the pattern in the series where elements are formed by repeating digits like 1, 22, 333,... up to nine 9s. For options A, B, and C, we must find expressions that relate the sum of the series at stage \( n \) and \( n-1 \).

Step 2: Finding a General Term

The general term of the series for an \( n \)-digit number is formed by repeating digit \( n \), \( n \) times. Thus, it can be expressed as \( n \times 111\ldots n\text{ times} \). The number \( 111\ldots \) with \( n \) digits can be expressed as \( \frac{10^{n}-1}{9} \). Therefore, each term is \( n \times \frac{10^{n}-1}{9} \).

Step 3: Calculating a Single Term Difference

To find the difference \( S_n - S_{n-1} \), we need to calculate the effect of adding the \( n^{th} \) term. This term is \( n \times \frac{10^{n} - 1}{9} \). So, \( S_n - S_{n-1} = n \times \frac{10^{n} - 1}{9} \).

Step 4: Checking Option C

Option C states \( 9(S_n - S_{n-1}) = n(10^n - 1) \). Substituting our expression for \( S_n - S_{n-1} \), we get \( 9\left(n \times \frac{10^{n} - 1}{9}\right) = n(10^n - 1) \). This confirms the expression matches, verifying option C.

Step 5: Verifying Calculations

Double-check the algebraic manipulation to ensure there are no errors in calculating \( S_n - S_{n-1} \). Observing both sides of C gives consistency as \( n \times (10^n - 1) = n \times (10^n - 1) \) when simplified.

Key Concepts

Sum of Series

A mathematical series is a sum of terms following a certain pattern or sequence. In the given exercise, the series consists of terms like 1, 22, 333, ..., and so on, up to 9 repeated nines. Understanding the sum of such a series requires recognizing the pattern in the numbers.
Each term can be thought of as a number made by repeating the digit multiple times, and they grow in magnitude as we progress through the series. Calculating the sum up to the nth term, denoted as \(S_n\), means finding the cumulative total from the first to the nth term.
The general formula for a term in our series is derived by noticing that each number can be expressed as \(n \times \frac{10^n - 1}{9}\). This formula comes from how the repeated digit structure can be mathematically represented, specifically using powers of 10 to capture the repeated sequence.

• The series starts with a simple single digit: 1.
• Progresses to two repeated digits: 22.
• Continues to three repeated digits: 333.

Understanding this sum helps us determine how each successive term builds on the previous one.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and expressions to simplify them or to find unknown values. This is crucial in dealing with mathematical series as seen in the exercise for determining the differences and sums of series at various stages.
For example, let's examine how we derived the relationship \(S_n - S_{n-1}\). This expression highlights the contribution of each additional term in the series when moving from the \((n-1)\)th term to the nth term.
The term for \(n\) can be broken down as follows:

• The number itself is expressed simply, when written as \(n \times \frac{10^n - 1}{9}\).
• Through algebraic manipulation, specifically scaling and subtraction, we find differences such as \(S_n - S_{n-1}\).

In step 4, the student's work involved verifying that multiplying by 9 simplifies the proposed expression, confirming the correctness of option C. This encourages confidence in the examination of series through algebraic techniques, honing one's problem-solving skills.

Mathematical Induction

Mathematical induction is a powerful method to prove that a statement is true for all natural numbers. Though not explicitly used in the solution presented, it’s a technique that helps confirm formulas or statements for larger series.
Though the exercise doesn't require induction, understanding its role illuminates how one might prove the validity of the sum formulas across a similar or extended series.

• Base Case: Start by proving the statement is true for a small value, like \(n = 1\).
• Inductive Step: Assuming it’s true for \(n = k\), prove it for \(n = k+1\).

Imagine applying induction here: once you understand how to derive \(S_n\) for smaller values manually, induction helps extrapolate this pattern to indefinite size. Such methods build foundational skills in both rigorous proof and logical reasoning in mathematics.