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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.

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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! $=\sum _{r=1}^{n}r.r!\phantom{\rule{0ex}{0ex}}=\sum _{r=1}^{n}\left[\left(r+1\right)-1\right]r!\phantom{\rule{0ex}{0ex}}=\sum _{r=1}^{n}\left[\left(r+1\right)r!-r!\right]\phantom{\rule{0ex}{0ex}}=\sum _{r=1}^{n}\left\{\left(r+1\right)!-r!\right\}\phantom{\rule{0ex}{0ex}}=\left(2!-1!\right)+\left(3!-2!\right)+...\left[\left(n+1\right)!-n!\right]\phantom{\rule{0ex}{0ex}}=\left[\left(n+1\right)!-1!\right]\phantom{\rule{0ex}{0ex}}=\left[\left(n+1\right)!-1\right]=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{proved}.$

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