Problem 10 - A Complete Resource Book in Mathematics for JEE Main

Question

The statement \(p(n): 1 \times 1 !+2 \times 2 !+3 \times 3 !+\ldots\) \(+n \times n !=(n+1) !-1\) is (A) true for all \(n>1\) (B) not true for any \(n\) (C) true for all \(n \in N\) (D) None of these

Step-by-Step Solution

Answer

(C) true for all \( n \in N \).

Step 1: Understand the Problem

We're given a statement involving factorials and sums: \( p(n): 1 \times 1! + 2 \times 2! + 3 \times 3! + \ldots + n \times n! = (n+1)! - 1 \). We must determine if this equation is true for certain values of \(n\), or not true at all.

Step 2: Verify for Initial Values

First, let's test the statement for \( n = 1 \). Calculate both sides:The left side: \( 1 \times 1! = 1 \).The right side: \( (1+1)! - 1 = 2! - 1 = 2 - 1 = 1 \).Both sides are equal, so it holds for \( n = 1 \).

Step 3: Inductive Hypothesis

Assume the statement is true for \( n = k \). That is, assume:\[ 1 \times 1! + 2 \times 2! + \ldots + k \times k! = (k+1)! - 1 \].

Step 4: Induction Step

We need to prove the statement is true for \( n = k+1 \). Consider:\[ 1 \times 1! + 2 \times 2! + \ldots + k \times k! + (k+1)(k+1)! \].By the inductive hypothesis:\[ = (k+1)! - 1 + (k+1)(k+1)! \].Simplify the right side:\[ (k+1)! + (k+1)(k+1)! - 1 = (k+1)! + (k+2)(k+1)! - 1 \].This can be rewritten as:\[ (k+2)! - 1 \].So it holds true for \( n = k+1 \).

Step 5: Conclusion

Since the statement is true for \( n = 1 \) and true for \( n = k+1 \) whenever it is true for \( n = k \), by the principle of mathematical induction, the statement is true for all natural numbers \( n \).

Key Concepts

Factorial

When we talk about factorial, we are dealing with an operation that is foundational in combinatorics, analysis, and many areas of mathematics. The symbol for factorial is an exclamation mark - for example, "n!". This notation means the product of all positive integers from 1 up to the number n. For example:

• 1! = 1
• 2! = 2 × 1 = 2
• 3! = 3 × 2 × 1 = 6

In general, for a natural number n, it means multiplying n by the factorial of (n-1). Factorials are particularly useful in calculating permutations and combinations, where the order of objects is significant. So factorials lay the groundwork for many proofs and problems involving sequences and series, including the one in this exercise.

Natural Numbers

Natural numbers are the counting numbers starting from 1 and progressing onwards as long as you keep adding. These numbers are

• 1, 2, 3, 4, ...

They do not include zero, negative numbers, or fractions. In many mathematical problems and proofs, natural numbers serve as the domain, especially when talking about sequences and series.
Natural numbers are foundational to the concept of mathematical induction. This method of proof is primarily concerned with showing propositions hold for every natural number, as seen in the given problem about factorial sums.

Inductive Hypothesis

The inductive hypothesis is a central step in the method of mathematical induction, which is a powerful tool for proving statements about natural numbers. It's like claiming a base assumption that the truth of a statement at a certain stage implies the truth of it at the next stage.
When you assume the statement to be true for n = k, you write:- \(1 \times 1! + 2 \times 2! + \ldots + k \times k! = (k+1)! - 1\)By using this assumption, you attempt to prove that:- \(1 \times 1! + 2 \times 2! + \ldots + (k+1) \times (k+1)! = (k+2)! - 1\)Once this is proven, the basis (that it's true for some starting number, usually n = 1) and the inductive step together confirm the statement for all natural numbers, linking beautifully with the principle of mathematical induction.