P(n) is the statement 10n + 3 × 4n + 2 + 5 is ÷ by 9
P(1) = 101 + 3 × 42 + 5 = 10 + 3 × 16 + 5
= 10 + 48 + 5 = 63 ÷ by 9
So P(1) is true. Assume that P(k) is true
(i.e.) 10k + 3 × 4k + 2 + 5 is ÷ by 9
(i.e.) 10k + 3 × 4k + 2 + 5 = 9C (where C is an integer)
⇒ 10k = 9C – 5 – 3 × 4k + 2 ….. (1)
To prove P(k + 1) is true.
Now P(k + 1) = 10k + 1 + 3 × 4k + 3 + 5
= 10 × 10k + 3 × 4 × 4k + 2 + 5
= 10[9C – 5 – 3 × 4k + 2] + 3 × 4k + 2 × 4 + 5
= 10[9C – 5 – 3 × 4k + 2] + 12 × 4k + 2 + 5
= 90C – 50 – 30 × 4k + 2 + 12 × 4k + 2 + 5
= 90C – 45 – 18 × 4k + 2
= 9[10C – 5 – 2 × 4k + 2] which is ÷ by 9
So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.