Question

The value of \( {}^2P_1 + {}^3P_1 + \dots + {}^nP_1 \) is equal to:

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

Hint:

If you forget the formula, test with $n=2$. The sum is just ${}^2P_1 = 2$. Plugging $n=2$ into option (E) gives $(4+2-2)/2 = 3

Answer 1

Answer: (E)

Concept: The permutation formula is \( {}^nP_r = \frac{n!}{(n-r)!} \). For \( r = 1 \), this simplifies significantly: \( {}^nP_1 = n \).

Step 1: Simplify each term in the series.
\[ {}^2P_1 = 2 \] \[ {}^3P_1 = 3 \] \[ \dots \] \[ {}^nP_1 = n \] The series is \( 2 + 3 + 4 + \dots + n \).

Step 2: Use the sum of an arithmetic progression.
The sum of natural numbers from \( 1 \) to \( n \) is \( \frac{n(n+1)}{2} \). Our series starts from 2, so we subtract 1 from the total sum: \[ \text{Sum} = \frac{n(n+1)}{2} - 1 \]

Step 3: Combine the terms.
\[ \frac{n^2 + n}{2} - \frac{2}{2} = \frac{n^2 + n - 2}{2} \]