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

Prove that: 1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + … + n . P(n, n) = P(n + 1, n + 1) – 1.

Prove that: 1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + … + n . P(n, n) = P(n + 1, n + 1) – 1.

Solution:

By making use of the formula,

$P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)!$

$P\text{ }\left( n,\text{ }n \right)\text{ }=\text{ }n!/\left( n-n \right)!$

$=\text{ }n!/0!$

$=\text{ }n!\text{ }\left[ Since,\text{ }0!\text{ }=\text{ }1 \right]$

Let’s take the LHS:

$=\text{ }1.\text{ }P\left( 1,\text{ }1 \right)\text{ }+\text{ }2.\text{ }P\left( 2,\text{ }2 \right)\text{ }+\text{ }3.\text{ }P\left( 3,\text{ }3 \right)\text{ }+\text{ }\ldots \text{ }+\text{ }n\text{ }.\text{ }P\left( n,\text{ }n \right)$

$=\text{ }1.1!\text{ }+\text{ }2.2!\text{ }+\text{ }3.3!\text{ }+\ldots \ldots \ldots +\text{ }n.n!\text{ }\left[ Since,~P\left( n,\text{ }n \right)\text{ }=\text{ }n! \right]$

$=\sum\limits_{r=1}^{n}{r.r!}$

$=\sum\limits_{r=1}^{n}{r.r!}+r!-r!$

$=\sum\limits_{r=1}^{n}{(r+1)r!-r!}$

$=\sum\limits_{r=1}^{n}{(r+1)-r!}$

$=\text{ }\left( 2!-1! \right)\text{ }+\text{ }\left( 3!-2! \right)\text{ }+\text{ }\left( 4!-3! \right)\text{ }+\text{ }\ldots \ldots \ldots \text{ }+\text{ }\left( n!-\left( n-1 \right)! \right)\text{ }+\text{ }\left( \left( n+1 \right)!-n! \right)$