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Question -

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



Answer -

By using the formula,

P (n, r) = n!/(n – r)!

P (n, n) = n!/(n – n)!

= n!/0!

= n! [Since, 0! = 1]

Consider LHS:

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

= 1.1! + 2.2! + 3.3!+………+ n.n! [Since, P(n, n) = n!]

= (2! – 1!) + (3! –2!) + (4! – 3!) + ……… + (n! – (n – 1)!) + ((n+1)! – n!)

= 2! – 1! + 3! – 2! +4! – 3! + ……… + n! – (n – 1)! + (n+1)! – n!

= (n + 1)! – 1!

= (n + 1)! – 1[Since, P (n, n) = n!]

= P(n+1, n+1) – 1

= RHS

Hence Proved.

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