Question

The value of \(\frac{C(n,2){(n+1)!}\) is}

• A. \(\frac{1}{2}e + 1\)
• B. \(e + 1\)
• C. \(\frac{1}{2}e - 1\)
• D. \(e\)

Hint:

Convert factorial expressions to match \(e^x\) series.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves summation of combinations combined with factorials, which typically points toward the exponential series \( e \).
: Key Formula or Approach:
Identity: \( \frac{{}^nC_2}{(n+1)!} = \frac{n(n-1)}{2(n+1)!} \).
Step 2: Detailed Explanation:
The general term is \( T_n = \frac{n(n-1)}{2(n+1)!} \). For \( n<2 \), \( T_n = 0 \). So the sum starts from \( n=2 \). \[ S = \frac{1}{2} \sum_{n=2}^{\infty} \frac{n(n-1)}{(n+1)n(n-1)(n-2)!} \] Actually, it's easier to use: \[ \frac{n(n-1)}{(n+1)!} = \frac{n^2-n}{(n+1)!} = \frac{(n^2+n) - 2n}{(n+1)!} = \frac{n(n+1) - 2(n+1) + 2}{(n+1)!} \] \[ = \frac{n(n+1)}{(n+1)!} - \frac{2(n+1)}{(n+1)!} + \frac{2}{(n+1)!} \] \[ = \frac{1}{(n-1)!} - \frac{2}{n!} + \frac{2}{(n+1)!} \] Now sum from \( n=2 \) to \( \infty \): \[ S = \frac{1}{2} \left[ \sum_{n=2}^{\infty} \frac{1}{(n-1)!} - 2 \sum_{n=2}^{\infty} \frac{1}{n!} + 2 \sum_{n=2}^{\infty} \frac{1}{(n+1)!} \right] \] 1. \( \sum_{n=2}^{\infty} \frac{1}{(n-1)!} = \frac{1}{1!} + \frac{1}{2!} + \dots = e - 1 \).
2. \( \sum_{n=2}^{\infty} \frac{1}{n!} = \frac{1}{2!} + \frac{1}{3!} + \dots = e - 1 - 1 = e - 2 \).
3. \( \sum_{n=2}^{\infty} \frac{1}{(n+1)!} = \frac{1}{3!} + \frac{1}{4!} + \dots = e - 1 - 1 - 1/2 = e - 2.5 \).
Substitute: \[ S = \frac{1}{2} [ (e-1) - 2(e-2) + 2(e-2.5) ] \] \[ S = \frac{1}{2} [ e - 1 - 2e + 4 + 2e - 5 ] \] \[ S = \frac{1}{2} [ e - 2 ] = \frac{e}{2} - 1 \].
Step 3: Final Answer:
The sum is \( \frac{1}{2}e - 1 \).