Prove that: $1. P (1, 1) + 2. P (2, 2) + 3. P (3, 3) + . . . + n . P (n, n) = P (n + 1, n + 1) - 1$

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Solution

$L H S = {1.}^{1} P_{1} + {2.}^{2} P_{2} + {3.}^{3} P_{3} + . . . . . . + n^{n} P_{n}$
As $^{n} P n = n!$ , hence
$L H S = 1.1! + 2.2! + 3.3! + . . . . + n . n!$
$= (1.1! + 1) + 2.2! + 3.3! + . . . . . + n . n! - 1$
$= 2! + 2.2! + 3.3! + . . . . + n . n! - 1$
$= 2! (1 + 2) + 3.3! + . . . . + n . n! - 1$
$= 3.2! + 3.3! + . . . . + n . n! - 1$
$= 3! + 3.3! + . . . . . . + n . n! - 1$
$= 3! (3 + 1)_{.} . . . . + n . n! - 1$
$= 4.3! + . . . . . + n . n! - 1$
$= 4! + . . . . . + n . n! - 1$
Similarly LHS will reduce to
$L H S = n! + n . n! - 1$
$= n! (n + 1) - 1$
$= (n + 1)! - 1$
$\Rightarrow L H S =^{n + 1} P_{n + 1} - 1 = R H S$

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