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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.
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!
\[= \sum^n_{r = 1} r . r!\]
\[ = \sum^n_{r = 1} \left[ \left( r + 1 \right) - 1 \right] r!\]
\[ = \sum^n_{r = 1} \left[ \left( r + 1 \right) r! - r! \right]\]
\[ = \sum^n_{r = 1} {(r + 1)! - r!} \]
\[ = \left( 2! - 1! \right) + \left( 3! - 2! \right) + . . . \left[ \left( n + 1 \right)! - n! \right]\]
\[ = \left[ \left( n + 1 \right)! - 1! \right] \]
\[ = \left[ \left( n + 1 \right)! - 1 \right] = \text{RHS}\]
\[ \text{Hence, proved} .\]
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