﻿
Question
Sun May 08, 2011

# Using the principle of mathematical induction prove that for all n belongs to N: 11 power"n+2" + 12 power"2n+1" is divisible by 133

Sun May 08, 2011
PART 1:
First, prove it for the base case of n = 1:

11^(1+2) + 12^(2+1)
= 11^3 + 12^3
= 1331 + 1728
= 133(10) + 1 + 1728
= 133(10) + 1729
= 133(10) + 133(13)
= 133(26)

PART 2:
Assume it is true for a natural number k. Prove it is true for the number k+1:

True for k:
11^(k+2) + 12^(2k+1) = 133(m), where m is an integer.

For k+1:
11^(k+1+2) + 12^(2(k+1)+1)
= 11^(k+2) * 11 + 12^(2k + 2 + 1)
= 11^(k+2) * 11 + 12^(2k + 1) * 12²
= 11 * 11^(k+2) + (11 + 133) * 12^(2k+1)
= 11( 11^(k+2) + 12^(2k+1) ) + 133 * 12^(2k+1)
= 11( 133m ) + 133 * 12^(2k+1)
= 133 ( 11m + 12^(2k + 1))

The stuff in the parentheses will be an integer, so the result is a multiple of 133.

Therefore by induction a number of that form is divisible by 133.

Related Questions
Fri September 30, 2016

# [(25)^2/3]^3/4 = ?

Sun September 11, 2016

﻿