# Is ts solution correct, I’m not understanding the break down from the beginning. Please explain Transcribed Image Text: Inductive step: Let P (k) be true, thus 7k – 2k is divisible

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2 (5m) = 5 [74 – 2m] Since 7k – 2m is in the set of all, we have 5 divides 7k+1 – 2k+1. Thus, P (k + 1) is true. Hence, 7″ – 2″ is divisible by 5, for each integer n > 0. |

Is ts solution correct, I’m not understanding the break down from the beginning. Please explain Transcribed Image Text: Inductive step:

Let P (k) be true, thus 7k – 2k is divisible by 8 and by the definition

of divisible, there exists an integer m such that 7k – 2k = 5m.

We need to prove that P (k + 1) is true.

7k+1 – 2k+1 = 7 · 7k – 2 · 2k

–

= 5- 7k + 2· 7k – 2 · 2k

%3D

= 5 – 7k + 2 [7* – 2*]

%3D

= 5- 7k + 2 (5m)

= 5 [74 – 2m]

Since 7k

– 2m is in the set of all, we have 5 divides 7k+1 – 2k+1.

Thus, P (k + 1) is true.

Hence, 7″ – 2″ is divisible by 5, for each integer n > 0.

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