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Let P(n) = 1×1! + 2×2! + 3×3! +…+ n×n

P(n): 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all n ϵ N

Step1:

P(1) = 1×1! = (2)! – 1 = 1

Thus, P(n) is equal to (n + 1)! – 1 for n = 1

Step2:

Let, P(m) be equal to (m + 1)! – 1

Then, 1×1! + 2×2! + 3×3! +…+ m×m! = (m + 1)! – 1

Now, we need to show that P(m+1) is true whenever P(m) is true.

P(m+1) = 1×1! + 2×2! + 3×3! +…+ m×m! + (m+1)×(m+1)!

= (m+1)! – 1 + (m+1)×(m+1)!

= (m+1)!(m+1+1) – 1

= (m+1)!(m+2) – 1

= (m+2)! – 1

Thus, P(m+1) is true.

So, by the principle of mathematical induction, P(n) is true for all nϵN.

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