Question

using the principle of mathematical induction prove that (2n+7)<(n+3)^2

Answer 1

Let P(n): (2n + 7) < (n + 3)² 

When = 1, LHS = (2 × 1 + 7) = 9 and RHS = (1 + 3)² = 4² = 16. 

Clearly, 9 < 16. 

Thus, P(n) is true for n = 1, i.e., P(1) is true. 

Let P(k) be true. Then

P(k): (2k + 7) < (k + 3)². ... (i) 

Now, 2(k + 1) + 7 = (2k + 7) + 2

                              < (k + 3)² + 2 = (k² + 6k + 11) [using(i)] 

                              < (k² + 8k + 16) = (k + 4)² = (k + 1 + 3)².

P(k + 1): 2(k + 1) + 7 < (k + 1 + 3)². 

P(k + 1) is true, whenever P(k) is true. 

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. 

Hence, by Principle of Mathematical Induction, P(n) is true for all n ∈ N.
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