prove that `2nlt(n+2)!` for all natural numbers n.

1.	prove that `2nlt(n+2)!` for all natural numbers n.
Answer» Consider the statement `P(n):2nlt(n+2)!` for all natural number n. Step I We observe that, P(1) is true `P(1):2(1)lt(1+2)!` `rArr 2lt3!rArr2lt3xx2xx1rArr2lt6` Hence, P(1) is true. Step II Now, assume that p(n) is true for n=k, `P(k):2klt(k+2)!` is true Step III Now, assume that P(n) is true for n=k, `P(k):2klt(k+2)!` is true. Step III To prove P(k+1) is true, we have to show that `P(k+1):2(k+1)lt(k+1+2)!` Now, `2klt(k+2)!` `2k+2lt(k+2)!` `2k+2ltIk+2)!+2` `2(k+1)lt(k+2)!+2` . . .(i) Also, `(k+2)!+2lt(k+3)!` . . . (ii) From Eqs. (i) and (ii), `2(k+1)lt(k+1+2)!` So, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction P(n) is true.

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