For any natural number n, 7n – 2n is divisible by 5.
According to the question,
P(n) = 7n – 2n is divisible by 5.
So, substituting different values for n, we get,
P(0) = 70 – 20 = 0 Which is divisible by 5.
P(1) = 71 – 21 = 5 Which is divisible by 5.
P(2) = 72 – 22 = 45 Which is divisible by 5.
P(3) = 73 – 23 = 335 Which is divisible by 5.
Let P(k) = 7k – 2k be divisible by 5
So, we get,
⇒ 7k – 2k = 5x.
Now, we also get that,
⇒ P(k+1)= 7k+1 – 2k+1
= (5 + 2)7k – 2(2k)
= 5(7k) + 2 (7k – 2k)
= 5(7k) + 2 (5x) Which is divisible by 5.
⇒ P(k+1) is true when P(k) is true.
Therefore, by Mathematical Induction,
P(n) = 7n – 2n is divisible by 5 is true for each natural number n.
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