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:

1.P (1, 1) + 2. P (2, 2) + 3. P (3, 3) + ... + n . P (n, n) = P (n + 1, n + 1) − 1

P (n,n) = n!

1.1! + 2.2! + 3.3! ......+ n.n! = (n+1)! − 1

LHS = 1.1! + 2.2! + 3.3! ......+ n.n!

$=\sum_{r=1}^{n} r . r !$

 

$=\sum_{r=1}^{n}[(r+1)-1] r !$

$=\sum_{r=1}^{n}[(r+1) r !-r !]$

 

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

$=(2 !-1 !)+(3 !-2 !)+\ldots[(n+1) !-n !]$

$=[(n+1) !-1 !]$

 

$=[(n+1) !-1]=\operatorname{RHS}$

Hence proved.