If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 1

1.	If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all nϵ N.
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 nϵ 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.

If P(n) is the statement “n(n + 1) is even”, then what is P(3)? 1×1! + 2×2! + 3×3! +…+ n×n! = (n + 1)! – 1 for all nϵ N.

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 nϵ 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.