00:01
Okay, so for part a of problem 8, we have an.
00:06
We're given in our problem that an is 4n minus 2.
00:14
Now, so if we substitute n by m minus 1, so we have n, a .m minus 1 is 4 times n minus 1 minus 2.
00:26
So, an minus am minus 1 should be 4n minus 2 minus 4 n minus 2 minus 4 times n minus 1 plus 2.
00:42
And the result is 4.
00:45
So, an can be found by taking the sum of a .m minus 1 and 4, where n is equal to 2, 3, and dot, dot.
01:02
And also have to notify that a1 is 2.
01:07
Okay, our b.
01:10
Now we have an is equal to 1 plus negative 1 to the power of n.
01:17
Now if we substitute n by n minus 1, then we have 1 plus negative 1 to the power of n minus 1.
01:28
Now we take the difference, a n minus 1, which is equal to 1 plus negative 1 to negative 1 to the n minus 1 minus 1 to the negative 1 to the power of n minus 1.
01:46
And the result is 2 times negative 1 to the power of n.
01:54
So that our a .n can be found by 2 times negative 1 to the power of n plus a .m minus 1...