The sequence an $n$  0 is defined by a0 = 1, a1 = – 4 and an + 2 = – 4an + 1 – 7an, for $n$  0 . Find the number of positive integer divisors of a50 – a49a51.

Step 1: The sequence {an}, $n$  0 and a0 = 1, a1 = – 4 and given that an + 2 = –4an + 1 – 7an ….(1)

Step 2: Now an2 – an + 1. an –1 = an2 – ( 4an – 7an –1 ) an –1

Step 3: = an2 + 4an . an –1 + 7an2–1

Step 4: = –an ( 7an –2 ) + 7an2 –1

Step 5: = 7 an2–1 – an .an –2 )

Step 6: = 72 an2–1 – an –1.an –3 )

Step 7: => ( )

Step 8: an2 – an + 1 an –1 = 7n –1 a12 – a0a2 = 7n –1(16 – 1.9) = 7n

Step 9: a50 – a49 .a51 = 750

Step 10: => Number of positive integer divisors of 750 = 50 + 1 = 51