The value of P 1 2 + P 1 3 + ? + P 1 n is equal to

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Solution

Q.
The value of ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$ is equal to

1 $\frac{n^{2} - n + 2}{2}$

2 $\frac{n^{2} + n + 2}{2}$

3 $\frac{n^{2} + n - 1}{2}$

4 $\frac{n^{2} + n - 2}{2}$

Ans.
(4)

We have, ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$

$= 2 + 3 + \dots + n = \frac{n (n + 1)}{2} - 1 = \frac{n^{2} + n - 2}{2}$

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