CBSE Class 9 Answered
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
Asked by Theodosia Lourdes | 08 May, 2011, 04:04: AM
Expert Answer
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.
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.
Answered by | 08 May, 2011, 12:41: PM
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