Question

41n-14n is a multiple of 27 prove the following by using the principle of mathematical induction

Answer 1

Answer:

41n-14n is a multiple of 27

Step-by-step explanation:

$\text{Let P(n): 41n-14n is a multiple of 27}$

$\text{put n=1,}$

$\text{P(1): 41-14=27 which is a multiple of 27}$

$\therefore\text{P(1) is true}$

$\text{Assume that P(k) is true}$

$\text{That is, 41k-14k is multiple of 27}$

$\implies\:41k-14k=27c$ where c is an integer

$\text{To prove: P(k+1) is true}$

$\text{that is to prove: 41(k+1)-14(k+1) is a multiple of 27}$

$\text{Now,}$

$41(k+1)-14(k+1)$

$=41k+41-14k-14$

$=(41k-14k)+41-14$

$=27c+27$

$=(c+1)27\:\text{which is a multiple f 27}$

$\implies\:\text{P(k+1) is true}$

$\text{Hence, P(n) is true for all natural numbers}$

Answer 2

Step-by-step explanation:

this anwer is give here clearly

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