Show that 133 divides 11n+2 + 122n+1 for every natural number n.
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Explanation:
First, prove it for the base case ofn=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)
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.
Fork+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)×12
2
=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)
) hence it is divisible by 133.
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