Question:medium

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

Show 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
Updated On: May 6, 2026
  • \( \frac{n^2 - n + 2}{2} \)
  • \( \frac{n^2 + n + 2}{2} \)
  • \( \frac{n^2 + n - 1}{2} \)
  • \( \frac{n^2 - n - 1}{2} \)
  • \( \frac{n^2 + n - 2}{2} \)
Show Solution

The Correct Option is

Solution and Explanation

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