Q7 of 30 Page 70

For any natural number n, 7n – 2n is divisible by 5.

Given; P(n) = 7n – 2n is divisible by 5.

P(0) = 70 – 20 = 0; is divisible by 5.


P(1) = 71 – 21 = 5; is divisible by 5.


P(2) = 72 – 22 = 45; is divisible by 5.


P(3) = 73 – 23 = 335; is divisible by 5.


Let P(k) = 7k – 2k is divisible by 5; ⇒ 7k – 2k = 5x.


⇒ P(k+1) = 7k+1 – 2k+1


= (5 + 2)7k – 2(2k)


= 5(7k) + 2 (7k – 2k)


= 5(7k) + 2 (5x); is divisible by 5.


⇒ P(k+1) is true when P(k) is true.


∴ By Mathematical Induction P(n) = 7n – 2n is divisible by 5 is true for each natural number n.


More from this chapter

All 30 →
5

n3 – 7n + 3 is divisible by 3, for all natural numbers n.

 6

32n – 1 is divisible by 8, for all natural numbers n.

 8

For any natural number n, xn – yn is divisible by x – y, where x integers with x ≠ y.

 9

n3 – n is divisible by 6, for each natural number n.