Question

If n(p) = 1x1! + 2x2! + 3x3! + ... nxn! = (n+1) xn! then n is

• A. n $\in$ N
• B. n is an even integer
• C. no n exits

Answer 1

Answer: (C)

(iii) no n exists

Answer 2

The problem asks for which values of 'n' the given equality holds:

$1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + n \times n! = (n+1) \times n!$

Let's evaluate the sum on the left-hand side (LHS). The general term of the sum is $k \times k!$. We can rewrite $k \times k!$ as follows:

$k \times k! = (k+1-1) \times k! = (k+1)k! - 1 \times k! = (k+1)! - k!$

Now, let's substitute this into the sum: LHS $= \sum_{k=1}^{n} k \times k! = \sum_{k=1}^{n} ((k+1)! - k!)$

This is a telescoping sum: LHS $= (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((n+1)! - n!)$ Notice that most terms cancel out: LHS $= (n+1)! - 1!$ LHS $= (n+1)! - 1$

Now, let's look at the right-hand side (RHS) of the given equality: RHS $= (n+1) \times n!$ By the definition of factorial, $(n+1)! = (n+1) \times n!$. So, RHS $= (n+1)!$

Now, we equate the LHS and RHS as per the given condition: $(n+1)! - 1 = (n+1)!$

To solve for 'n', subtract $(n+1)!$ from both sides of the equation: $-1 = 0$

This is a mathematical contradiction. The statement $-1 = 0$ is false. This implies that the original equality $1 \times 1! + 2 \times 2! + \dots + n \times n! = (n+1) \times n!$ is never true for any value of 'n'.

Therefore, no value of 'n' exists for which the given statement holds.