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

Asked by Aaryan | 1 year ago |  40

#### 1 Answer

##### Solution :-

=$$\dfrac{n!}{0!}$$

= 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.

Answered by Sakshi | 1 year ago

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