Answer :

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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