00:03
We are given the first term and the recurrence formula for a sequence, and we're told that this sequence converges.
00:13
And we're asked to find the first term of the sequence is negative 1, and the recurrence relation is am plus 1, as an plus 6 over am plus 2.
00:28
As we're told of this sequence converges, we have the limit as an approach as infinity of am plus 1 exists, and is equal to some real number l.
00:38
This is equal to the limit as n approaches infinity, a n plus 6 over a .m plus 2.
00:50
And by limit rules, this is equal to the limit as end approaches infinity of a .m.
00:59
Plus 6 over the limit as n approaches infinity on a .m.
01:06
Plus 2.
01:08
And this is equal to l plus 6 over l plus 6 over l plus for that little number of l.
01:19
And so we have the l is equal to l plus 6 of l plus 2.
01:22
You can multiply those sides by l plus 2.
01:24
We get l squared plus 2l equals l plus 6.
01:30
So we have that l squared plus l minus 6 is equal to 0.
01:36
This can be factored as l minus 2 times l plus 3 equals 0...